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I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is weighted homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every weighted homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a weighted homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = $$ $$e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) by Saito is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is weighted homogeneous.

Saito, K.
Quasihomogene isolierte Singularitäten von Hyperflächen.
Invent. Math 14, 123–142 (1971).
https://doi.org/10.1007/BF01405360

I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is weighted homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every weighted homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a weighted homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = $$ $$e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth projective hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) by Saito is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is weighted homogeneous.

Saito, K.
Quasihomogene isolierte Singularitäten von Hyperflächen.
Invent. Math 14, 123–142 (1971).
https://doi.org/10.1007/BF01405360

I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is weighted homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every weighted homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a weighted homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) by Saito is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is weighted homogeneous.

I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is weighted homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every weighted homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a weighted homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = $$ $$e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) by Saito is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is weighted homogeneous.

Saito, K.
Quasihomogene isolierte Singularitäten von Hyperflächen.
Invent. Math 14, 123–142 (1971).
https://doi.org/10.1007/BF01405360

I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is quasi-homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every quasi-homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a quasi-homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) conjectured by Saito and proved by Steven S.-T. Yau and Huaiqing Zuo is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is quasi-homogeneous.

I am just posting my comment as one answer. For every complex polynomial $f(z_1,\dots,z_n)$, denote by $f^*$ the polynomial map, $$f^*:\mathbb{C}^n\setminus \text{Zero}(f) \to \mathbb{C}\setminus \{0\}.$$ A complex polynomial $f(z_1,\dots,z_n)$ is weighted homogeneous of degree $d$ if there exists an ordered $n$-tuple of positive integers $(e_1,\dots,e_n)$ such that for every monomial $z_1^{d_1}\cdots z_n^{d_n}$ whose coefficient in $f$ is nonzero, we have $d_1e_1 + \dots + d_ne_n$ equals $d$, i.e., every nonzero term of $f$ has degree $d$ when each variable $z_i$ has degree $e_i$.

Proposition. For every weighted homogeneous polynomial $f(z_1,\dots,z_n)$ of degree $d$, if there exists a divisor $m>1$ of $d$ such that $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a weighted homogeneous polynomial $g(z_1,\dots,z_n)$ of degree $d/m$, then every fiber of $f^*$ is disconnected. Otherwise, every fiber of $f^*$ is connected, and even irreducible and smooth.

Proof. For every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, consider the homogenization of the fiber of $f^*$ over $\lambda$, i.e., inside the weighted projective space $\mathbb{CP}(1,e_1,\dots,e_n)$ with variables $(w,z_1,\dots,z_n)$, denote by $X_\lambda$ the hypersurface with defining equation $\lambda w^n - f(z_1,\dots,z_n) = 0$. The fiber of $f^*$ over $\lambda$ is canonically isomorphic to the affine open $X_\lambda \cap D_+(w)$, where $D_+(w)$ is the open subset of the weighted projective space where $w$ is nonzero.

Consider $w^n-f(z_1,\dots,z_n)$ as a polynomial in $w$ with coefficients in $\mathbb{C}(z_1,\dots,z_n)$. Since we know the splitting field and Galois extension of the polynomial $w^n-z$ over $\mathbb{C}(z)$, the polynomial $w^n-f(z_1,\dots,z_n)$ factors if and only if $f(z_1,\dots,z_n)$ equals $g(z_1,\dots,z_n)^m$ for a divisor $m>1$ of $n$ and an element $g(z_1,\dots,z_n)$ of $\mathbb{C}(z_1,\dots,z_n)$. In that case, by Gauss's Lemma, also $g(z_1,\dots,z_n)$ is a quasi-homogeneous polynomial of degree $d/m$. In this case, the fiber of $f^*$ over $\lambda = \mu^m$ equals the disjoint union of the fibers of $g^*$ over $\zeta\mu$ for all $m^{\text{th}}$ roots of unity $\zeta$. So every fiber is disconnected.

Conversely, if $\lambda w^m-f(z_1,\dots,z_n)$ is irreducible for every element $\lambda$ of $\mathbb{C}\setminus\{0\}$, then the hypersurface $X_\lambda$ is irreducible, and hence the open affine $X_\lambda \cap D_+(w)$ is also irreducible. QED

In the case of interest to the author of the original post, the polynomial $f$ is even homogeneous, namely of the form $L_1\cdots L_d$ for an ordered $d$-tuple of pairwise linearly independent linear polynomials $L_i(z_1,\dots,z_n)$. Thus, every fiber is irreducible.

Remark. Another way to see smoothness / submersivity of $f^*$ is to observe that the associated polynomial $h(y_1,\dots,y_n) = f(y_1^{e_1},\dots,y_n^{e_n})$ is homogeneous in the usual sense, hence satisfies the Euler identity. $$d\cdot h(y_1,\dots,y_n) = e_1\cdot y_1^{e_1-1}\frac{\partial f}{\partial z_1}(y_1^{e_1},\dots,y_n^{e_n}) + \dots + e_n\cdot y_n^{e_n-1} \frac{\partial f}{\partial z_n}(y_1^{e_1},\dots,y_n^{e_n}).$$ Thus, at every point where every partial derivative $\partial f/\partial z_i$ vanishes, then also $d\cdot h(y_1,\dots,y_n)$ vanishes, and hence the corresponding point $(z_1,\dots,z_n) = (y_1^{e_1},\dots,y_n^{e_n})$ is in $\text{Zero}(f)$.

In particular, if $\text{Zero}(f)$ is a smooth hypersurface, then the unique critical point of the polynomial map $f$ is at $(0,\dots,0)$. A beautiful theorem (of Oda?) by Saito is that the converse also holds: for every polynomial map $f$ from $\mathbb{C}^n$ to $\mathbb{C}$ such that the unique critical point is $(0,\dots,0)$, mapping to $0$, there exists a set of polynomial coordinates $(z_1,\dots,z_n)$ of $\mathbb{C}^n$ (i.e., a polynomial automorphism of the original coordinates) such that $f$ is weighted homogeneous.