We claim that only $Q = \pm P^k$ works for any positive $k$, WLOG the initial coefficient of $P,Q$ be positive (else if replace by $-P,-Q$) so for large enough $x$, $P(x)$ and $Q(x)$ both are positive, now fix $x$ and use the following well known result, Result: If for infinitely many $n$ $a^n-1 \mid b^n-1$ then $b$ is a power of $a$. Now notice since these are polynomials with bounded degree, there must be some value(as power) that occurs infinitely times, say $k$ then $Q = P^k$

This is a solid overkill, the result used requires very strong theorems to be proven.

Since we have $a^{2n}-1 \mid b^{2n}-1$, we can write this as $(a^2)^n-1 \mid (b^2)^n-1$ now the result is somewhat easy to prove, and is well known,(for eg. see here)