Let $K$ be a field. Let $\mathfrak{g}$ be a Lie algebra over $K$. Let $\lambda:\mathfrak{g}\to K$ be a homomorphism. Let $V_\lambda$ be $K$ endowed with the structure of right $\mathfrak{g}$-module given by $\lambda$.

For a representation $V$ of $\mathfrak{g}$, let us say that $\lambda$ is a **weight** of $V$ is $V_\lambda$ is isomorphic to a subquotient of $V$ as $\mathfrak{g}$-module. This means that for some block-triangulation of the representation, $\lambda$ appears as a diagonal block. Say that $\lambda$ is a **strong weight** if $V_\lambda$ is isomorphic to a submodule of $V$, i.e., if there exists $y\in V\smallsetminus\{0\}$ such that $x\dot y=\lambda(x)y$ for all $x\in\mathfrak{g}$. For $\mathfrak{g}$ nilpotent and $\dim(V)<\infty$, weight and strong weight are equivalent, but not in general: for instance, for the adjoint representation of the 2-dimensional Lie algebra $\mathfrak{b}$ with basis $u,v$, $[u,v]=v$ and $\lambda(u)=0$ and $\lambda(v)=0$, the weights are $0$ and $\lambda$, but only $\lambda$ is a strong weight.

By definition, we have $H_1(\mathfrak{g},V_\lambda)=\mathrm{Ker}(\lambda)/I_\lambda$, where $I_\lambda\subset\mathfrak{g}$ is the image of $d:\Lambda^2\mathfrak{g}\to\mathfrak{g}$, $x\wedge y\mapsto \lambda(x)y-\lambda(y)x-[x,y]$.

If $\lambda=0$, then $H_1(\mathfrak{g},V_0)$ is the abelianization, and hence is zero iff $\mathfrak{g}$ is perfect (which is the same as $0$ being a strong weight of the coadjoint representation. Note that for $\mathfrak{g}$ finite-dimensional, $0$ is a weight of the adjoint/coadjoint representation iff it is not semisimple (for $\mathfrak{g}$ solvable this means iff $\mathfrak{g}\neq 0$). It is a strong weight of the adjoint representation iff $\mathfrak{g}$ has nontrivial center.

Next I assume $\lambda\neq 0$.

The non-vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that there exists a linear form $f$ on $\mathfrak{g}$, not proportional to $\lambda$, such that $f([x,y])=\lambda(x)y-\lambda(y)x$ for all $x,y$. The latter condition means that $x\mapsto \lambda(x)u+f(x)v$ is a homomorphism into the 2-dimensional Lie algebra $\mathfrak{b}$, and the non-proportionality means that it is nonzero. In other words, the vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that every lift $\mathfrak{g}\to\mathfrak{b}$ of $\lambda$ has a 1-dimensional image.

It is clear that the latter condition only depends on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

Also it's clear that if $\mathfrak{g}$ is nilpotent then $\mathfrak{b}$ is not quotient of $\mathfrak{g}$, so in this case $H_1(\mathfrak{g},V_\lambda)=0$ for all $\lambda\neq 0$.

For $\mathfrak{g}$ finite-dimensional metabelian and $\lambda\neq 0$, let me check that
$H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is weight of the adjoint representation, if and only if $\lambda$ is a strong weight of the adjoint representation.

Proof: the latter equivalence is true for metabelian Lie algebras and $\lambda\neq 0$. Indeed, if $\mathfrak{a}$ is a Cartan subalgebra and $\mathfrak{n}$ is the derived subalgebra, let $\lambda$ be a nonzero weight of $\mathfrak{g}$. So it is a strong weight of $\mathfrak{n}$ on $\mathfrak{n}$. If $y\neq 0$ and $[x,y]=\lambda(x)y$ for all $\in\mathfrak{a}$, then this is also true for $x\in\mathfrak{n}$ (because $\mathfrak{n}$ is abelian), and hence for all $x$ (since $\mathfrak{a}+\mathfrak{n}=\mathfrak{g}$).

If the $H_1(\mathfrak{g},V_\lambda)$ is nonzero, we have a surjection $\mathfrak{g}\to\mathfrak{b}$ projecting to $\lambda$ and by pull-back we deduce that $\lambda$ is a weight of $\mathfrak{g}$.

Conversely, $\lambda$ is a weight of $\mathfrak{g}$, by the above it is a strong weight. So there exists $y$ such that for all $x\in\mathfrak{g}$ we have $[x,y]=\lambda(x)y$. Since $\lambda\neq 0$, necessarily $y\in\mathfrak{g}^{(1)}$, the derived subalgebra, and this can be viewed as a condition on the adjoint representation of $\mathfrak{g}/\mathfrak{g}^{(1)}$ on $\mathfrak{g}^{(1)}$. We can decompose this representation into isotypic components, and kill all other components (for other eigenvalues as $\lambda$), and also kill an invariant hyperplane in the isotypic component of $\lambda$. After doing this, we preserve the property of $\lambda$ being a weight (maybe $y$ is not longer the same), and in addition the derived subalgebra is 1-dimensional. Using a Cartan subalgebra, we see that $\mathfrak{g}$ is then semidirect product of an abelian subalgebra $\mathfrak{a}$ with the 1-dimensional $\mathfrak{g}$, and killing the kernel of $\lambda$ in $\mathfrak{a}$ results in a 2-dimensional algebra. This quotient is precisely a 2-dimensional surjective lift of $\lambda$ onto $\mathfrak{b}$.

**Corollary:** for $\mathfrak{g}$ finite-dimensional and $\lambda\neq0$

$H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is a (strong) weight of the adjoint representation of $\mathfrak{g}$ on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

(a) Now here's an example for which $\lambda$ is a weight of the adjoint representation of $\mathfrak{g}$, but not of its metabelianization (so $H_1(\mathfrak{g},V_\lambda)=0$ anyway). Namely, consider a Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=x$, $[s,y]=y$, $[s,z]=2z$, $[x,y]=z$. (This appears a parabolic subalgebra in $\mathfrak{su}(2,1)$ over the reals.) Define $\lambda$ mapping $s$ to 1 and other basis elements to 0. Then $2\lambda$ is a (strong) weight of the adjoint representation (with eigenspace generated by $z$), but not a weight of the 3-dimensional metabelianization (which only has the weights $0$ and $\lambda$), so $H_1(\mathfrak{g},V_{2\lambda})=0$. [Edit: actually $H_*(\mathfrak{g},V_{2\lambda})=0$, see (e) below.]

Edits: here are remarks about $H_0$ and $H_d$ for $d=\dim(\mathfrak{g})$. First, we have $H_0(\mathfrak{g},V_\lambda)=0$ iff $\lambda\neq 0$.

(b) For $\mathfrak{g}$ finite-dimensional, say of dimension $d$, define $\tau_{\mathfrak{g}}(x)=\mathrm{Trace}(\mathrm{ad}(x))$ (for instance, $\mathfrak{g}$ is unimodular iff $\tau_{\mathfrak{g}}= 0$). Then $H_d(\mathfrak{g},V_\lambda)=0$ if and only if $\lambda=\tau_{\mathfrak{g}}$. However, $\tau_{\mathfrak{g}}$ is often not a weight.

(c) For instance, if $\mathfrak{g}$ is 3-dimensional with 1-dimensional abelianization and weights of the adjoint representation $0$, $\lambda$ and $t\lambda$ with $t\notin\{0,-1\}$, then $\tau_{\mathfrak{g}}=(1+t)\lambda$ is not a weight of the adjoint representation although $H_*(\mathfrak{g},V_{\tau_{\mathfrak{g}}})\neq 0$.

(d) Looking at $H_2$ also provides unimodular counterexamples. Namely, consider an $(n+1)$-dimensional Lie algebra with basis $(s,v_1,\dots,v_n)$ with $[s,v_i]=a_iv_i$. (It is unimodular iff $\sum a_i=0$.) Define $\lambda(s)=1$, $\lambda(v_i)=0$, so the weights are $0$ and the $a_i\lambda$ (which are strong weights). For $i\neq j$ we have $H_2(\mathfrak{g},V_{(a_i+a_j)\lambda})\neq 0$, although usually $(a_i+a_j)\lambda$ is not a weight; this gives unimodular counterexamples to the conjecture in dimension $n+1\ge 4$.

(e) Variant of (a): consider the Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=ax$, $[s,y]=by$, $[s,z]=(a+b)z$, $[x,y]=z$, with $a,b,c$ scalars with $a,b,a+b\neq 0$. Let $\lambda$ map $s$ to $1$ and other basis elements to $0$. So the weights are the $t\lambda$ for $t\in\{0,a,b,a+b\}$.

Then a direct computation shows that the homology of $V_{t\lambda}$ is nonzero exactly for $t\in\{0,a,b,2a+b,a+2b,2a+2b\}$ (for $0$ the non-vanishing occurs in degree $0, 1$; for $a,b$ it occurs in degree $1$ and $2$; for $2a+b$ and $a+2b$ it occurs in degree $2$ and $3$ and for $2a+2b$ it occurs in degree $4$).

Thus in this example **both implications** of the conjecture fail: $(a+b)\lambda$ is a weight of the adjoint representation but $V_{(a+b)\lambda}$ has zero homology, while $t\lambda$ for $t\in\{2a+b,a+2b,2a+2b\}$ are not weights of the adjoint representation but for these values $V_{t\lambda}$ has homology.

Cohomologie des groupes topologiques et des algèbres de Lie" might contain relevant information. Unfortunately it's not easy to find. For instance it probably includes the fact that $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ nilpotent and all $\lambda\neq 0$. $\endgroup$ – YCor May 7 at 10:51