If $deg(P)=0$, then $P(x)\equiv 0$ or $P(x)\equiv-1$. If $deg(P)>0$, then $P(z)=a\prod_{j}(z-z_{j})\Longrightarrow a=1$. If $z$ is root P(x), then $z^{2}$ is root too. Therefore all roots are $exp(\frac{2\pi i k_{j}}{m})$. If $z^{2}$ root, but $z=exp(\frac{\pi ik_{j}}{m})$ is not root we have $P(z)=-2$. It give solutions $P(x)=x^{m}-1$.

Assume $P(x)\not \equiv-1$ and set $Q(X): =P(X)+1$. Then (*) $Q(x^{2})=Q(x)^{2}$ and $Q(x)\not\equiv 0$. Take $n\in\mathbb{N}_{0}$ maximal with $x^{n}|Q(x)$ and write $Q(x)=x^{n}R(x)$ with $R(x)\in\mathbb{C}[x],\ R(0)\not =0$. Then (*) gives (**) $R(x^{2})=R(x)^{2}$. If $R(x)\equiv c\not =0$ constant, then (**) gives $c=c^{2}$ and $c=1$. This gives $P(x)=x^{n}-1$ with $n\in\mathbb{N}_{0}$. ($\equiv 0$ for $n=0$) So assume $\deg R\geq 1$, and then take $m\in\mathbb{N}_{0}$ maximal with $R(x)=S(x^{2^{m}})$. ($2^{m}\leq \deg R$) Then (**) gives $S(x^{2^{m+1}})=S(x^{2^{m}})^{2}$ and $S(x^{2})-S(x)^{2}$ has an infinite number of roots. So (***) $S(x^{2})=S(x)^{2}$ and $S(-x)=S((-x)^{2})=S(x^{2})=S(x)^{2}$, which gives $S(-x)=S(x)$ or $S(-x)=-S(x)$. But $S(0)=R(0)\not =0$, so $S(-x)\not =-S(x)$. And so $S$ is even and $S(x)=T(x^{2})$ for some $T\in\mathbb{C}[x]$, which contradicts the choice of $m$.

A good solution , Olorin , congratulation But i think , the problem should be in $ \mathbb{R}[x]$