Constant $P$ and linear $P$ do not work, henceforth we assume $\deg P > 1$. Obviously $m = \deg P^2$. Since this holds for all integers $x$ it holds for all complex numbers as a polynomial identity. Consider the polynomial $Q(x) = P(x)-x$. If $z$ is a root of $Q$, then $P(P(z)) = P(z) = z$ and it follows then, that $z^m+2 = 0$. Therefore, all roots of $Q$ are also roots of $x^m+2$. As a result, there exists some $k$ such that $Q(x) \mid (x^m+2)^k$. Since $Q$ is in $\mathbb{Z}[x]$ and $x^m+2$ is irreducible over $\mathbb{Q}[x]$ we have $Q(x) = c(x^m+2)^k$ for some $k \geq 2$ and $c \in \mathbb{Z}$. But this is a contradiction since $\deg Q \leq \deg P < \deg P^2$.