We known $\forall P\in \mathbb C(x)=\{\dfrac N D:(N, D) \in \mathbb C[x] \} $, if $P(\mathbb Z) \subset \mathbb Z$ then $P\in \mathbb Q[x] $.
Is it true that $\forall P\in \mathbb C(x, y)=\{\dfrac N D:(N, D) \in \mathbb C[x, y] \} ,$
if $P(\mathbb Z, \mathbb Z)=\{P(n,m) : (n,m)\in\mathbb Z^2\} \subset \mathbb Z$ then $ P\in \mathbb Q[x, y] $?
This is well known, but I'll give a short proof using 3 dimensions as an example.
Every power $x^n$ can be written as the integer linear combination of binomial coefficients $\binom xj$ for $0\le j\le n$. So we can write our polynomial as a finite sum $$p(x,y,z)=\sum_{i,j,k} a_{i,j,k} \binom xi\binom yj\binom zk.$$ Now consider the lexicographically least $(i,j,k)$ such that $x^iy^jz^k$ is present in $p$. Then $p(i,j,k)=a_{i,j,k}$ so $a_{i,j,k}$ is an integer. Now subtract off $a_{i,j,k}\binom xi\binom yj\binom zk$ and repeat. Finally we have that all $a_{i,j,k}$ are integers.
