Let $Q(x) = P(x) - 1$, so $Q(x^2) = Q(x)^2$. $Q = 0$ is a solution, so assume $Q \neq 0$. Also, if $x \mid Q$, then we can divide out the highest power of $x$ dividing $Q$ to get a new solution which does not have $0$ as a root, so let us also assume that $Q$ does not have $0$ as a root. The equation implies that if $r$ is a root of $x$, so is $r^2$. Thus if there exists a root $r$ of $Q$ with magnitude not equal to $1$, there exist infinitely many roots of $Q$, contradiction. Thus every root of of $Q$ has magnitude $1$. The equation also implies that if $e^{i\theta}$ is a root of $Q$, so is $e^{i\theta/2}$. Thus if $e^{i\theta}$ is a root of $Q$ with $0 < \theta < 2\pi$, $Q$ again has infinitely many roots, contradiction. Finally, if $1$ is a root of $Q$, so is $-1$, again a contradiction. Thus $Q$ has no roots and is constant. Because $Q(x^2) = Q(x)^2$ and $Q \neq 0$, it must be $1$. Undoing all of our transformation and assumptions, the solutions are $P(x) = 1$ and $P(x) = 1 + x^k$ for $k \ge 0$.

If $P(x)=c \to P(x)=1,2$ Let $P(x)=ax^n+Q(x), deg(Q(x))<n$ $ax^{2n}+Q(x^2)+2ax^n+2Q(x)=a^2x^{2n}+Q(x)^2+2ax^nQ(x)+2 \to a=1, deg(Q(x))=0$ $P(x)=x^n+c$ $x^{2n}+c+2x^n+2c=x^{2n}+c^2+2cx^n+2 \to c=1$ So $P(x)=x^n+1$ Answer: $P(x)=1,P(x)=2,P(x)=x^n+1,n \geq 1$