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 <b>quasi-homogeneous</b> 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 $e_i$.  

<B>Proposition</B>.  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.

<B>Proof</B>.  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.  <B>QED</B>

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 if irreducible.