The constant polynomial $P(x) = c$ is a solution if and only if $c = c^2$, thus the polynomials $P(x) = 0$ and $P(x) = 1$ are solutions of the problem.
We claim that the only polynomial of a positive degree $n$ which solves the equation is of the form $P(x) = (x-1)^n$. In view of the identity $(x^2 - 1)^n = (x-1)^n(x+1)^n$, the latter is clearly a solution for any $n \ge 1$.
If $a x^n$ ($a \neq 0$) is the leading term of a polynomial $P(x)$ of a positive degree $n$, then $a x^{2n}$ is the leading term of the polynomial $P(x^2)$ and $a^2 x^{2n}$ is the leading term of the polynomial $P(x) P(x+2)$. If $P$ satisfies the given equality, comparing the leading order terms thus gives $a = a^2$, hence $a = 1$. The polynomial $P$ can therefore be written in the form $P(x) = (x-1)^n + Q(x)$, where $Q$ is either identically zero, or is a nonzero polynomial of degree $k$, where $0 \le k < n$. Comparing the polynomials
$$P(x^2) = (x^2 - 1)^n + Q(x^2), \\ P(x) P(x+2) = [(x-1)^n + Q(x)][(x+1)^n + Q(x+2)]$$
we obtain (upon multiplying out the brackets and cancelling the terms $(x^2 - 1)^n$ on both sides) the equality
$$Q(x^2) = (x-1)^n Q(x+2) + (x+1)^n Q(x) + Q(x) Q(x+2).$$
The zero polynomial $Q$ clearly satisfies this relation. For a nonzero $Q$ of degree $k < n$, however, $Q(x^2)$ is a polynomial of degree $2k$, while on the right-hand side of the last equation there is a polynomial of degree $n+k$ (whose leading term is $2b x^{n+k}$, if $b x^k$ is the leading order term of the polynomial $Q(x)$). Since $2k < n+k$, this is not possible.
Conclusion. The solutions are the constant polynomials $P(x) = 0$ and $P(x) = 1$ and the polynomial $P(x) = (x-1)^n$ for any natural number $n$.