Suppose that $A$ is a Noetherian commutative ring such that: (1) the nilradical (intersection of all prime ideals) of $A$ vanishes, and (2) localization at every maximal ideal is a finite ring. Prove that $A$ is finite.
Here is my work so far: Suppose on the contrary that $A$ is infinite. Then we may find infinitely many distinct elements $a_1, a_2, a_3, \dots$ in $A$. Since $A$ is Noetherian, the chain of ideals $(a_1) \subseteq (a_1, a_2) \subseteq (a_1, a_2, a_3) \subseteq \cdots$ have a maximal ideal, say $\mathfrak m := (a_1,\dots, a_N)$. Then $\mathfrak m = (a_1,a_2,\dots)$. Let $x \in A - \mathfrak m$. Then since $\mathfrak m$ is a maximal ideal, it must be that $A = (a_1,\dots,a_N,x)$. Let $D := A - \mathfrak m$. Then the residue field $D^{-1}A$ is finite. I am not sure where to go from here. I think that there may be a way to use the fact that the nilradical of $A$ is the intersection of all prime ideals of $A$, but I don't see how.
Let $A$ be a reduced Noetherian ring such that its localization at every maximal ideal is finite. This implies that the localization at every prime ideal is finite: for any prime ideal $\mathfrak{p}$, we can pick a maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$. This would imply that $A_{\mathfrak{p}} = (A_{\mathfrak{m}})_{\mathfrak{p}}$. The localization of any finite ring is finite since there are finitely many choices for the numerator and denominator of a valid fraction.
Next, for any prime ideal $\mathfrak{p}$, the kernel of the map $A \to A_{\mathfrak{p}}$ is the ideal of elements annihilated by some element of $A - \mathfrak{p}$. Consider the set of ideals $\{\mathrm{Ann}(\alpha): \alpha \in S - \mathfrak{p}\}$. For any $\mathrm{Ann}(\alpha_1)$ and $\mathrm{Ann}(\alpha_2)$ in this set, we know that $\mathrm{Ann}(\alpha_1\alpha_2)$ also lies in the set and contains both ideals. Thus, the set contains at most one maximal element. We also know that $A$ is Noetherian, so it contains at least one maximal element. Finally, the sum of all the ideals in the set is the kernel. It follows that there exists some $\alpha_{\mathfrak{p}} \in S - \mathfrak{p}$ such that the kernel is $\mathrm{Ann}(\alpha_{\mathfrak{p}})$.
Finally, observe that since $A_{\mathfrak{p}}$ is finite, $A/\mathrm{Ann}({\alpha_{\mathfrak{p}}})$ is finite for each $\mathfrak{p}$. If we assume that $A$ is infinite, this implies that for any finite collection of prime ideals $\mathfrak{p}_1, \mathfrak{p}_2, \ldots, \mathfrak{p}_k$, the intersection $\bigcap_{i=1}^k \mathrm{Ann}(\alpha_{\mathfrak{p}_i})$ is finite index and consequently non-empty. In other words, $\mathrm{Ann}(\alpha_{\mathfrak{p}_1}, \ldots, \alpha_{\mathfrak{p}_k})$ is non-zero. Once again, since $A$ is Noetherian, there exist finitely many primes $\mathfrak{p}_1, \ldots, \mathfrak{p}_k$ such that $(\alpha_{\mathfrak{p}_1}, \ldots, \alpha_{\mathfrak{p}_k}) = (\alpha_{\mathfrak{p}}: \text{$\mathfrak{p}$ is prime})$. Thus, there exists a non-zero element $f$ which annihilates every $\alpha_{\mathfrak{p}}$. Since $\alpha_{\mathfrak{p}} \not \in \mathfrak{p}$, this implies that $f \in \mathfrak{p}$. However, this must hold true for all primes $\mathfrak{p}$, which contradicts the fact that $A$ is reduced. Thus, $A$ is finite.