This is true of a ring $R$ iff $\operatorname{Spec} R$ is a finite discrete space (or equivalently, the quotient of $R$ by its nilradical is a finite product of fields).
First, suppose $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$ for every $R$-module $M$. We will first prove that every prime ideal in $R$ is maximal. Suppose $\mathfrak{p}\subset R$ is prime but not maximal. Let $S=R/\mathfrak{p}$; then $S$ is a domain but not a field. Let $a\in S$ be a nonzero element which is not a unit, and consider the $R$-module $M=\bigoplus_{n\in\mathbb{N}} S/(a^n)$. Let $I=\bigcap_n (a^n)\subset S$ and let $J$ be the inverse image of $I$ in $R$. It is clear that $I$ is the annihilator of $M$ as an $S$-module, so $J$ is the annihilator of $M$ as an $R$-module.
On the other hand, $I$ generates a proper ideal in the localization $S[a^{-1}]$. Indeed, since $a$ is not a zero divisor, the only way $I$ could fail to generate a proper ideal is if $a^n\in I$ for some $n$, which would imply $a^n\in (a^{n+1})$ so $a^n=a^{n+1}b$ for some $b\in S$. Again since $a$ is not a zero divisor, this implies $1=ab$ so $a$ is a unit, contrary to our choice of $a$.
Thus $I$ generates a proper ideal in $S[a^{-1}]$, and we can extend it to a maximal ideal. This maximal ideal pulls back to a prime ideal $\mathfrak{q}\subset R$ such that $J\subseteq\mathfrak{q}$ but $\bar{a}\not\in\mathfrak{q}$ where $\bar{a}\in R$ is an element whose image in $S$ is $a$. Since every element of $M$ is annihilated by some power of $\bar{a}$, $M_\mathfrak{q}=0$, so $\mathfrak{q}\not\in \operatorname{Supp}(M)$. But since $J\subseteq \mathfrak{q}$, $\mathfrak{q}\in V(J) =V(\operatorname{Ann M})$. This contradicts our assumption that $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$ for all $R$-modules $M$.
Thus every prime in $R$ is maximal. Now if $\mathfrak{p}$ is any maximal ideal, the module $R/\mathfrak{p}$ has support $\{\mathfrak{p}\}$. Thus for any subset $A\subseteq\operatorname{Spec} R$, the module $M=\bigoplus_{\mathfrak{p}\in A}R/\mathfrak{p}$ has support $A$. Since $\operatorname{Supp}(M)=V(\operatorname{Ann M})$ and $V(I)$ is a closed set for any ideal $I$, this implies every subset of $\operatorname{Spec} R$ is closed, so $\operatorname{Spec} R$ is discrete. Finally, since $\operatorname{Spec} R$ is quasicompact, this implies it is also finite.
Conversely, suppose $R$ is a ring such that $\operatorname{Spec} R$ is a finite discrete space. Then $R$ can be identified with the finite product $\prod_{\mathfrak{p}}R_\mathfrak{p}$ of its localizations at all its prime ideals. When we make this identification, each prime ideal $\mathfrak{p}$ corresponds to the elements of $\prod_{\mathfrak{p}}R_\mathfrak{p}$ whose $\mathfrak{p}$-coordinate is in the unique maximal ideal of $R_\mathfrak{p}$. Also, any $R$-module $M$ is naturally isomorphic to the direct sum $\bigoplus_{\mathfrak{p}}M_\mathfrak{p}$ of its localizations at all the prime ideals. The annihilator of $M$ is then the product $\prod_\mathfrak{p} \operatorname{Ann}_{R_\mathfrak{p}} M_\mathfrak{p}\subseteq\prod_{\mathfrak{p}}R_\mathfrak{p}$. Thus $\operatorname{Ann} M\subseteq\mathfrak{p}$ iff $\operatorname{Ann}_{R_\mathfrak{p}} M_\mathfrak{p}$ is contained in the unique maximal ideal of $R_\mathfrak{p}$. Since any proper ideal in $R_\mathfrak{p}$ is contained in the unique maximal ideal, this means $\operatorname{Ann} M\subseteq\mathfrak{p}$ iff $M_\mathfrak{p}\neq 0$. That is, $\mathfrak{p}\in V(\operatorname{Ann} M)$ iff $\mathfrak{p}\in \operatorname{Supp}(M)$, so $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$.
