Find all polynomials $P(x)$ such that $P(x^2)=P(x)^2$

Hint If $P \neq 0$, then $P$ has the form $$P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 ,$$ where $a_n \neq 0$. Comparing the leading terms of $P(x^2)$ and $P(x)^2$ gives that $a_n = 1$. If $P(x) \neq x^n$, then there is some largest index $m < n$ such that $a_m \neq 0$, and so $P$ has the form $$P(x) = x^n + a_m x^m + O(x^{m - 1}) .$$ Now, substitute in both sides of the condition.

Doing so gives $$P(x^2) = x^{2n} + a_m x^{2m} + O(x^{2m - 1})$$ and $$P(x)^2 = x^{2n} + 2 a_m x^{m + n} + O(x^{m + n - 1}) .$$ The second-largest nonzero term in $P(x^2)$ has degree $2m < m + n$, so comparing the degree $m + n$ terms gives $a_m = 0$, a contradiction. Thus, $P(x)$ must be $0$ or $x^n$ for some $n$. On the other hand, checking directly shows that these polynomials satisfy the condition.

Incidentally, this argument works over any field of characteristic $\neq 2$ (in characteristic $2$, $P(x^2) = P(x)^2$ holds for all polynomials).

Well, $P(0)$ must be $0$ or $1$, and as you note, if $P(0)=0$ then $P(x)=xQ(x)$ and it follows that $Q(x^2)=Q(x)^2$ as well.

Great so far. We can repeat this until we get a $Q$ that does not have $Q(0)=0$. So there exists $j\ge0$ and $Q$ so that $P(x)=x^jQ(x)$ and $$Q(x^2)=Q(x)^2,\quad Q(0)=1.$$

But there are problems with your argument about roots of unity. The relevant sequence is $r^{2^n}$, not $r^{2n}$. Yes, that sequence has to be periodic, but that does not imply that $r^{2n}=1$, it implies that $r^{2^j-2^k}=1$. The biggest problem I see is that having shown that each root is a root of unity of some order you jump to the conclusion that all roots of unity of that order are roots of $P$; I don't see why that would follow.

So. We assume that $Q(x^2)=Q(x)^2$ and that $Q(0)=1$.

If $Q\ne 1$ then $$Q(x)=1+a_kx^k+\dots,$$where $k>0$, $a_k\ne0$, and all the missing terms have order greater than $k$. Then $$Q(x^2)=1+a_kx^{2k}+\dots,$$while $$Q(x)^2=1+2a_kx^k+\dots.$$That's impossible unless $a_k=0$ and $k=0$, contradiction.

So $Q=1$; the only such polynomials are $P(x)=x^j$.