The original equation is equivalent to $P(x^2) - 1 = P^2(x) - 2P(x) + 1$, i.e.
$$P(x^2) - 1 = (P(x) - 1)^2.$$
If we define $Q(x) := P(x) - 1$, the above equation becomes
$$Q(x^2) = Q^2(x).$$
If $Q(x)$ is a constant polynomial, this implies that it has to be $0$ or $1$. On the other hand, if the degree of $Q$ is a positive integer $n$, we can write $Q(x) = a_n x^n + R(x)$, where $R$ is a polynomial of degree $r < n$. Plugging this into the equation, we get
$$a_n x^{2n} + R(x^2) = a_n^2 x^{2n} + 2a_n x^n R(x) + R^2(x).$$
We conclude $a_n = 1$ and $R(x^2) = 2a_n x^n R(x) + R^2(x)$.
Notice that the left-hand side of the above equality is of degree $2r$, while the right-hand side is of degree $n + r$. Since $r < n$, it follows that $r = 0$. Therefore, $Q(x) = x^n$.
The solutions of the initial equation are $P(x) = 1$, $P(x) = 2$ and $P(x) = x^n + 1$, for any positive integer $n$.