The solutions are $f(x) = 0$, $f(x) = 1$, $f(x) = \omega$ where $\omega$ is an $(m-1)$st root of unity, and $f(x) = x^m$.

If $f(x) = c$ is a constant polynomial, then the relation holds if and only if $c = c^m$.
Clearly, the solutions are $c = 0, 1$ and all the $(m-1)$st roots of unity.

If $f$ is a non-constant polynomial, then $f(x)$ attains infinitely many values.
Thus, there are infinitely many $y$ such that $f(y) = y^m$. Since $f$ is a polynomial,
this implies $f(x) = x^m$ for any $x$.

It is easy to check that all these are solutions.