Ext groups of modules over a noetherian ring (Hartshorne exercise III.6.7)

In Hartshorne exercise III.6.7, we are asked to show that if $X = \operatorname{Spec} A$ is an affine Noetherian scheme, and $M, N,$ $A$-modules with $M$ finitely generated, then $$ \operatorname{Ext}^i_A(M, N) \cong \operatorname{Ext}^i_X(\tilde{M}, \tilde{N}). $$ I'm wondering if the $M$ finitely generated hypothesis is even necessary. For since $A$ is Noetherian, sheafifying injective $A$-modules yields injective $\mathcal{O}_X$-modules, so that if $0 \to N \to I^\cdot$ is an injective resolution of $N$ as an $A$-module, one has $$ \operatorname{Ext}^i_A(M, N) = h^i( \operatorname{Hom}_A(M, I^\cdot)) \cong h^i(\operatorname{Hom}_{\mathcal{O}_X}(\tilde{M}, \tilde{I}^\cdot)) = \operatorname{Ext}^i_X(\tilde{M}, \tilde{N}). $$ Is there something wrong with this reasoning?

You're correct that $M$ finitely generated is not necessary for $\operatorname{Ext}$ (and the specifics of your solution to this part are correct), but it is necessary for $\mathcal{E}xt$, which is the second half of the problem.

To see that $M$ finitely generated is necessary for the second half of the problem, first observe that Hom only commutes with localization when the first argument is finitely presented. For instance, taking $A=N=\Bbb Z$ and $M=\Bbb Z[\frac12]$, $\operatorname{Hom}_A(M,N)=0$ but the Hom sheaf is nonzero over $D(2)\subset\operatorname{Spec} \Bbb Z$, so we get a counterexample involving $\operatorname{Ext}^0$.