Rewrite the given equation to

$$P\left(x^{2}\right)-1=(P(x)-1)^{2} \text {. }$$

Let $Q(x)=P(x)-1$, then $Q$ is a polynomial with real coefficients for which

$$Q\left(x^{2}\right)=Q(x)^{2}$$

Suppose $Q$ is constant, say $Q(x)=c$ with $c \in \mathbb{R}$. Then $c=c^{2}$, so $c=0$ or $c=1$. Both possibilities satisfy the equation. We can now assume that $Q$ is not constant, so we can write $Q(x)=b x^{n}+R(x)$ with $n \geq 1, b \neq 0$ and $R(x)$ a polynomial with real coefficients of degree at most $n-1$. The polynomial equation now becomes

$$b x^{2 n}+R\left(x^{2}\right)=b^{2} x^{2 n}+2 b x^{n} \cdot R(x)+R(x)^{2} .$$

By comparing the coefficients of $x^{2 n}$ on both sides, we get $b=b^{2}$. Since $b \neq 0$, it follows that $b=1$. If we now subtract $x^{2 n}$ from both sides, we find

$$R\left(x^{2}\right)=2 x^{n} \cdot R(x)+R(x)^{2} .$$

If $R$ is not the zero polynomial, then it has a degree $m \geq 0$. It holds that $m < 2m$. Contradiction. Therefore, $R$ must be the zero polynomial, which implies that $Q(x)=x^{n}$. This indeed satisfies the polynomial equation for $Q$.

This gives for $P$ the solutions $P(x)=1, P(x)=2$ and $P(x)=x^{n}+1$ with $n \geq 1$.