Let $R$ be a commutative ring, and let $I \subseteq R$ be a nilpotent ideal. Let moreover $M$ be an $R$-module, and let $IM$ be the submodule generated by the products $xm$ with $x \in I$ and $m \in M$. It is quite easy to prove that, if $M/IM$ is finitely generated as an $R$-module, then $M$ is finitely generated. Indeed, let $\tilde{m}_1, \ldots , \tilde{m}_k$ be generators of $M/IM$ and lift them to elements $m_1, \ldots, m_k$ in $M$. Call $N$ the submodule generated by $m_1, \ldots, m_k$: then, we see that $$ M= N + IM. $$ Indeed, take $m \in M$. By definition of the $m_i$, we have $r_1,\ldots,r_k \in R$ such that $m - \sum_i r_im_i$ lies in $IM$. Now, since $I$ is nilpotent, we conclude (Nakayamaa variant of Nakayama's lemma if you want) that $M=N$, and we are done.
What about finite presentation? Namely:
Let us assume that $I$ is nilpotent and also finitely presented as an $R$-module. Can I conclude that if $M/IM$ is finitely presented as an $R$-module, then the same is true for $M$? (if necessary, I can also assume that $R$ is a coherent ring).
