Let $R$ be a commutative ring, and let $I$ be an ideal. We assume that $I$ is nilpotent, so $I^n=0$ for some $n$. Moreover, we assume that it is finitely presented, namely it is the cokernel of some morphism of $R$-modules $R^n \to R^m$, in other words the following sequence is exact: $$ R^n \to R^m \to I \to 0. $$ Moreover, we assume that the quotient ring $R/I$ is coherent, namely, any finitely generated ideal is finitely presented (as an $R/I$-module).
Exercise 3.37 of this book asks to prove that, under the above assumptions, the ring $R$ is coherent.
My attempt goes on as follows. We take a finitely generated ideal $J$ of $R$ and we want to prove that it is finitely presented. So, we have a surjective $R$-homomorphism $$ \varphi \colon R^m \to J $$ We let $\widetilde{J}$ be the ideal of $R/I$ generated by the image of $$ R^m \xrightarrow{\varphi} J \hookrightarrow R \to R/I, $$ which can be easily seen (if I'm not mistaken) to be just $(J+I)/J$. This ideal is finitely generated, in particular the above composition factors through $(R/I)^m$ producing a surjective morphism $$ \overline{\varphi} \colon (R/I)^m \to (J+I)/I. $$ Since $R/I$ is assumed to be coherent, the kernel of $\overline{\varphi}$ is finitely generated. Now, I can quite easily check that the natural map $R^m \to (R/I)^m$ induces an $R$-linear morphism $$ \ker \varphi \to \ker \overline{\varphi}. $$ Then, I'm essentially stuck. I'd like to use nilpotency of $I$ to prove something like "$\ker \varphi / (I\ker \varphi)$ is finitely generated, so $\ker \varphi$ is finitely generated", but I can't see how to use the assumption that $I$ itself is finitely presented.