Let $R$ be a Noetherian domain of dimension two.
Let $\mathfrak{m}_1,\mathfrak{m}_2$ be two distinct maximal ideals of height two. Are there always infinitely many prime ideals contained in $\mathfrak{m}_1\cap\mathfrak{m}_2$?
I have done the following case:
Let $k$ be an algebraically closed field. Let $R=k[x,y]$. Let $\mathfrak{m}_1, \mathfrak{m}_2,\ldots,\mathfrak{m}_r$ be $r$ distinct maximal ideals. Then there are infinitely many prime ideals $\mathfrak{p}\subset \bigcap_{i=1}^r \mathfrak{m}_i$.
Proof: By Nullstellensatz, write $\mathfrak{m}_i=(x-a_i,y-b_i)$. If all $a_i$ are distinct. Then we may find a polynomial $F=y-f(x)$ which is irreducible (i.e. the interpolation polynomial). And we may add more points to get more different irreducible polynomials.
Since $\mathfrak{m}_i$ are distinct pairwise. We can always find a linear transformation $A\in GL_2$ so that we come back to above. Here it will use the field $k$ is infinite.
Corollary This is true for $\mathbb{A}^2_k$ for any field $k$.
It seems that it is difficult to go more far.
Let $A$ be a finitely generated $k$-algebra which is a domain of dimension two. Let $\mathfrak{m}_1\neq\mathfrak{m}_2$ be two maximal ideals. I am even unable to find a prime ideal $0\neq\mathfrak{p}\subset \mathfrak{m}_1\cap\mathfrak{m}_2$.
Thanks.