Clearly, $P(x) = x$ is one solution. Now we prove this is the only solution by showing $P(x)=x$ holds for an infinite number of integer $x$.

We do this by claiming all the integers defined by the series $a_{n+1}=a_n^2+1$, $a_0=0$, and prove $P(a_n)=a_n$ for all $n$ by induction.

Clearly because $P(0) =0$, we have $P(a_0)=a_0$. Also, $P(1)=P(0^2 + 1)=P(0)^2 + 1 = 1=a_1$. Assume $P(a_k)=a_k$ holds for all $k\ge 1$. Then $$P(a_{k+1})=P(a_k^2+1)=P(a_k)^2+1 = a_k^2+1 = a_{k+1}$$

Hence, by the principle of mathematic induction, we find $P(a_n)=a_n$ holds for all $n$.

Now, because $P(x)=x$ has an infinite number of distinct roots, this must be an identity.