By schur $P$ has infinitely many primes dividing it so let $q$ be a large enough prime dividing $P(n)$ for large enough $n$ then $q \mid P(r)$ where $n \equiv r \pmod q$ but if $r > N$ then $q \mid P(r) \mid r!$ so $q \le r \le q-1$ which is a contradiction, hence $r \in [0,N]$. By infinite pigeonhole some $r$ will repeat infinitely many times (for infinitely many such $q$) meaning $P(r)=0$. Note that $P$ cannot have a repeated root because if $(n-r)^2 \mid n!$ picking $n-r$ to be a large enough prime will give $\nu_{n-r}(n!) = 1$, a contradiction. Therefore $P$ has the form $(n-a_1)\cdots (n-a_k) Q(n)$ where $a_i$ are distinct non negative integers and $Q$ is irreducible. If $Q$ is not constant then we can repeat the same process as above to obtain a contradiction. Hence the solution set is \[P(n) = c\prod_{i=1}^{\deg P} (n - a_i), ~ a_i \neq a_j \in \mathbb Z_{\ge 0}, ~ c \in \mathbb Z_{>0}.\]