Let me provide an elementary proof that for $\mathfrak{g}$ nilpotent finite-dimensional and $\lambda\neq 0$ we have $H_*(\mathfrak{g},V_\lambda)=0$. (In general it seems to be particular case of results of Delorme in the 70s, and possibly known earlier.)

Recall that the homology of $V_\lambda$ is the homology of the Chevalley-Eilenberg complex, with $C_p=\Lambda^p\mathfrak{g}$, and $d_p:C_p\to C_{p-1}$ is given by $$x_1\wedge\dots\wedge x_p\mapsto \sum_{i=1}^p(-1)^{i+1}\lambda(x_i)x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge x_p+$$ $$+\sum_{1\le i<j\le p}[x_i,x_j]\wedge x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge \hat{x_j}\wedge\dots\wedge x_p.$$

Let $\mathfrak{z}(\mathfrak{g})$ be the center of $\mathfrak{g}$. Then for all $z\in\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$ and $u\in\Lambda^{p-1}\mathfrak{g}$ we have $d_p(u\wedge z)=d_{p-1}(u)\wedge z$ (because all remaining terms include $[x_i,z]$ or $\lambda(z)$ which are zero). The induction step is the following:

Lemma. Fix $p\ge 1$. Let $z$ be an element of $\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$. Suppose that the $H_i(\mathfrak{g}/Kz,V_\lambda)$ is zero for $i\in\{p-1,p\}$. Then $H_p(\mathfrak{g},V_\lambda)=0$.

Let $f$ be a cycle in degree $p$. Modulo $z$ it is a $p$-boundary. This means that there exists $g\in\Lambda^{p+1}\mathfrak{g}$ such that $d_{p+1}(g)=f$ in $\lambda^p\mathfrak{g}/Kz$, which means that $d_{p+1}(g)=f+\mu\wedge z$ for some $\mu\in\Lambda^{p-1}\mathfrak{g}$. Since $d_pf=0$, we moreover have $d_{p-1}(\mu)\wedge z=d_p(\mu\wedge z)=0$. This means that $d_{p-1}(\mu)=0$ in $\Lambda^{p-2}(\mathfrak{g}/Kz)$. Hence, there exists $\xi\in\Lambda^{p}\mathfrak{g}$ such that $d_{p}\xi=\mu$ in $\Lambda^{p-1}(\mathfrak{g}/Kz)$. This means that $d_p\xi=\mu+\eta\wedge z$ for some $\eta\in\Lambda^{p-2}\mathfrak{g}$. So $d_{p+1}(\xi\wedge z)=d_p(\xi)\wedge z=\mu\wedge z$ and hence $f$ is a boundary. (If $p=1$, we directly have $d_0\mu=0$ since $d_0=0$ so no need to introduce $\eta$.)

Corollary. $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ finite-dimensional nilpotent.

Given that $H_0(\mathfrak{g},V_\lambda)=0$ for every $\lambda\neq 0$, the lemma reduces, by induction, to the case when $\mathfrak{g}$ is abelian. In this case, we choose the basis $(e_0,\dots,e_n)$ with $\lambda(e_0)=1$, $\lambda(e_i)=1$ for $i\ge 1$, and view the Lie algebra as graded in $\mathbf{Z}^n$ with $e_i$ of grade $u_i$ (the canonical basis of $\mathbf{Z}^n$) and $e_0$ of grade $0$, noting that the boundary map preserves the grading. In $\Lambda^i\mathfrak{g}$, we have the grades $u_I=\sum_{i\in I}u_i$ when $I$ is a subset of $\{1,\dots,n\}$ of cardinal $i$ or $i-1$. If $I$ has cardinal $i$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by the $x_I=\wedge_{i\in I}e_i$, which is a $i$-cycle, and is also a boundary, namely of $e_0\wedge x_I$ (up to sign). If $I$ has cardinal $i-1$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by $y_I=e_0\wedge x_I$, which has a nonzero image by $d_i$. This proves the vanishing of all the homology.

Let me provide an elementary proof that for $\mathfrak{g}$ nilpotent finite-dimensional and $\lambda\neq 0$ we have $H_*(\mathfrak{g},V_\lambda)=0$. (In general it seems to be particular case of results of Delorme in the 70s, and possibly known earlier.)

Recall that the homology of $V_\lambda$ is the homology of the Chevalley-Eilenberg complex, with $C_p=\Lambda^p\mathfrak{g}$, and $d_p:C_p\to C_{p-1}$ is given by $$x_1\wedge\dots\wedge x_p\mapsto \sum_{i=1}^p(-1)^{i+1}\lambda(x_i)x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge x_p+$$ $$+\sum_{1\le i<j\le p}[x_i,x_j]\wedge x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge \hat{x_j}\wedge\dots\wedge x_p.$$

Let $\mathfrak{z}(\mathfrak{g})$ be the center of $\mathfrak{g}$. Then for all $z\in\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$ and $u\in\Lambda^{p-1}\mathfrak{g}$ we have $d_p(u\wedge z)=d_{p-1}(u)\wedge z$ (because all remaining terms include $[x_i,z]$ or $\lambda(z)$ which are zero). The induction step is the following:

Lemma. Fix $p\ge 1$. Let $z$ be an element of $\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$. Suppose that $H_i(\mathfrak{g}/Kz,V_\lambda)$ is zero for $i\in\{p-1,p\}$. Then $H_p(\mathfrak{g},V_\lambda)=0$.

Let $f$ be a cycle in degree $p$. Modulo $z$ it is a $p$-boundary. This means that there exists $g\in\Lambda^{p+1}\mathfrak{g}$ such that $d_{p+1}(g)=f$ in $\lambda^p\mathfrak{g}/Kz$, which means that $d_{p+1}(g)=f+\mu\wedge z$ for some $\mu\in\Lambda^{p-1}\mathfrak{g}$. Since $d_pf=0$, we moreover have $d_{p-1}(\mu)\wedge z=d_p(\mu\wedge z)=0$. This means that $d_{p-1}(\mu)=0$ in $\Lambda^{p-2}(\mathfrak{g}/Kz)$. Hence, there exists $\xi\in\Lambda^{p}\mathfrak{g}$ such that $d_{p}\xi=\mu$ in $\Lambda^{p-1}(\mathfrak{g}/Kz)$. This means that $d_p\xi=\mu+\eta\wedge z$ for some $\eta\in\Lambda^{p-2}\mathfrak{g}$. So $d_{p+1}(\xi\wedge z)=d_p(\xi)\wedge z=\mu\wedge z$ and hence $f$ is a boundary. (If $p=1$, we directly have $d_0\mu=0$ since $d_0=0$ so no need to introduce $\eta$.)

Corollary. $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ finite-dimensional nilpotent and $\lambda\neq 0$.

Given that $H_0(\mathfrak{g},V_\lambda)=0$ for every $\lambda\neq 0$, the lemma reduces, by induction, to the case when $\mathfrak{g}$ is abelian. In this case, we choose the basis $(e_0,\dots,e_n)$ with $\lambda(e_0)=1$, $\lambda(e_i)=1$ for $i\ge 1$, and view the Lie algebra as graded in $\mathbf{Z}^n$ with $e_i$ of grade $u_i$ (the canonical basis of $\mathbf{Z}^n$) and $e_0$ of grade $0$, noting that the boundary map preserves the grading. In $\Lambda^i\mathfrak{g}$, we have the grades $u_I=\sum_{i\in I}u_i$ when $I$ is a subset of $\{1,\dots,n\}$ of cardinal $i$ or $i-1$. If $I$ has cardinal $i$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by the $x_I=\wedge_{i\in I}e_i$, which is a $i$-cycle, and is also a boundary, namely of $e_0\wedge x_I$ (up to sign). If $I$ has cardinal $i-1$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by $y_I=e_0\wedge x_I$, which has a nonzero image by $d_i$. This proves the vanishing of all the homology.