Let $A$ be a ring and let $X=M{\rm Spec}(A) \subset{\rm Spec}(A)$ be the set of maximal ideals of $A$ with the induced Zariski topology. Can $X$ be an affine scheme? That is, can we find a ring $B$ such that $X$ is homeomorphic to ${\rm Spec}(B)$?
We know that this is true when $X$ is a closed subset, but what about in general? Can we find any resrtictions of $A$ such that $X$ would be an affine scheme?
There is a purely topological characterization of spaces which are homeomorphic to $\operatorname{Spec} R$ for some ring $R$. Such spaces are called spectral spaces.
One key property is that every irreducible closed subset has a unique generic point. In the maximal spectrum, all points are closed. In particular, if there exist irreducible closed subsets which are not singletons, then the maximal spectrum is not spectral. Whenever $\operatorname{Spec} A$ has irreducible closed subsets whose set of closed points is not a singleton, then the maximal spectrum will not be spectral. (Edited since originally I forgot that an irreducible closed subset could have a unique closed point e.g. if $A$ is local).
This is one obstruction to the maximal spectrum being spectral but not the only one, see Eric Wofsey's comment below.