Let $A$ be a commutative $\mathbb k$-algebra and let $X = \mathrm{Spec} A$ be the corresponding affine scheme.
For simplicity assume $\Bbbk$ is an algebraically closed field of characteristic $0$.
The stalk of the structure sheaf $\mathcal O_X$ at a point $x \in X$ can be defined as the direct limit $$ \mathcal O_{X, x} = \varinjlim_{U \ni x} \mathcal O_X (U) $$ over all open sets $U$ containing the point $x$, which is a local ring, i.e. with exactly one maximal ideal.
Out of curiosity, it seems that we can also define the "stalk" in two points $x, y$ by $$ \mathcal O_{X, x, y} = \varinjlim_{U \supset \{ x, y \}} \mathcal O_X (U). $$
Is it true that $\mathcal O_{X, x, y}$ has exactly two maximal ideals whenever $x$ and $y$ are two distinct closed points? It seems not too difficult to show it has at least two maximal ideals, because there are two surjective maps to $\Bbbk$, but I don't know how to show that there are no more maximal ideals.
More precisely, we have an injective map $\mathcal O_{X, x, y} \to \mathcal O_{X, x} \times \mathcal O_{X, y}$ and the latter has exactly two maximal ideals. But in general, injective algebra morphisms can "lose" maximal ideals (e.g. $\Bbbk [x] \hookrightarrow \Bbbk [[x]]$)...