Let $R$ be a noetherian ring, $I$ a nilpotent ideal in $R$, and $M$ a module over $R$, of infinite type, such that $M/IM = 0$. Is it necessarily the case that $M=0$?
If $M$ were of finite type, then we would immediately get a positive answer by Nakayama's Lemma. If $R$ is not noetherian, then I think I have a counterexample.
Yes, as Eric Wofsey points out, it is necessarily the case that $M=0$, and this is in fact true even if $R$ is not Noetherian! Since $I$ is nilpotent, let $n$ be such that $I^n=0$. Now, since $M\big/IM=0$, we have in other words that $M=IM$. By induction on $k$, we then have $M=I^kM$ for all $k$, whence in particular $M=I^nM=0$.
Perhaps instead, again as Eric points out, you are thinking of a nil ideal; this is an ideal in which every element is nilpotent. Now, if $R$ is Noetherian, then every nil ideal of $R$ is nilpotent. Indeed, if $I\leqslant R$ is any ideal, then, since $R$ is Noetherian, we have $I=\langle a_1,\dots,a_n\rangle$ for some $a_i\in R$. If $I$ is also nil, then for each $i\leqslant n$ there exists $k_i$ such that $a_i^{k_i}=0$. Let $k=\sum_{i=1}^nk_i$; can you show that $I^k=0$? In particular, by the paragraph above, the answer to your question is still yes even if "nilpotent" is replaced by "nil".
However, it is the case that the generalized statement does not hold in general if $R$ is not Noetherian. For example, let $R$ be the quotient $$\frac{\mathbb{Q}[x_n:n\in\mathbb{N}]}{\langle x_1^2,x_n-x_{n+1}^2:n\in\mathbb{N}\rangle}.$$ For convenience, denote the image of each $x_n$ in $R$ as $a_n$, and let $I$ be the ideal of $R$ generated by the $a_n$. We have $I^2=I$, ie $I\big/I^2=0$, since $a_n=a_{n+1}^2\in I^2$ for each $n\in\mathbb{N}$. Furthermore, by induction, we have $a_n^{2^{n}}=0$ for each $n\in\mathbb{N}$, so $I$ is generated by nilpotent elements and hence (why?) nil. But $I\neq 0$, so taking $M=I$ gives the desired counterexample.