With the polynomiality assumption a characterization is possible. I doubt that much can be said without it.

**Theorem:** Assuming that $F(x,y)$ is a polynomial in $x$ of degree $y$ for any $y\in \mathbb Z_{\geq 0}$ we have that
$$F(x,y)=\sum_{k\geq 0}\alpha_k (x+y)^k\binom{x+y-2k}{y-k}$$
for some arbitrary sequence $\alpha_0,\alpha_1,\dots.$

**Proof:** There exist coefficients $\alpha_{k,y}$ such that for each $y\in \mathbb Z_{\geq 0}$ we can write
$$F(x,y)=\sum_{0\le k\le y} \alpha_{k,y}(x+y)^k \binom{x+y-2k}{y-k}.$$
Notice that the terms in this expression are nonzero only if $k\le \min{x,y}$. We can prove by induction on $k$ that $\alpha_{k,y}$ doesn't depend on $y$. For the base case $k=0$ we have $F(x,0)=F(0,x)$ which gives $\alpha_{0,0}=\alpha_{0,x}$. For the general case, assume we have shown our claim for $k\le m-1$. Looking at $F(x,m)=F(m,x)$
for arbitrary $x\geq m$, we cancel out the equal terms that we have from the induction hypothesis and we are left with
$$\alpha_{m,m}(x+m)^m\binom{x-m}{0}=\alpha_{m,x}(m+x)^m\binom{x-m}{x-m}\implies \alpha_{m,m}=\alpha_{m,x}$$
finishing our proof.