Let $A$ be a ring s.t. $p$ is maximal ideal of $A$ and $\operatorname{Spec}(A)=\{p\}$. Suppose $p$ is made of nilpotent elements. Then $A$ is artinian.
$\textbf{Q:}$ How do I check $A$ is artinian here? If $A$ is artinian, certainly $p$ is f.g by $A$ artinian implying $A$ noetherian. Now $p$ is generated by nilpotents. Given an descending chain, $I_k$, it suffices to show $p^l$ vanishes identically for some large $l$. Then $p^l\subset\dots I_k\dots\subset p$ is decomposition series. This will force $I_k$ finite length by considering successive quotients. Thus it suffices to show $p$ is f.g. How do I check $p$ is f.g. here?
Ref: Manin, Introduction to the Theory of Schemes, Sec 1.3 (2) One-point Spectra on pg 14 of the book.
It seems you cannot really conclude anything else, and the result is false as stated. For example, consider $\mathbb{C}[x_1, x_2, ...]/(x_1^2, x_2^2, ...)$, a local ring with unique prime ideal $(x_1, x_2, ...)$ that is not finitely generated.