Suppose $f$ has degree $d$, and suppose that for some $p$, $f(p) = q^m$ is not a power of $p$. Observe that $f(p^{1 + k(q - 1)})$ is divisible by $q$ for all $k$ (by FLT) so \[f(p^{1 + k(q - 1)}) = q^{m(k)}\] for all $k$ and some $m(k)$. Now \[p^{(q - 1)d} = \lim_{k \rightarrow \infty}\left(\frac{f(p^{1 + (k + 1)(q - 1)})}{f(p^{1 + k(q - 1)})}\right) = \lim_{k \rightarrow \infty} {q^{m(k + 1) - m(k)}}\] where $p \ne q$. If $d > 0$ some fixed nontrivial power of $p$ must get arbitrary close to one of finitely many powers of $q$, none of which it can equal. This is absurd, so we conclude that $d = 0$. In this case, $f$ is constant and equal to a prime power; all these functions work. Otherwise, $f(p)$ is a power of $p$ for all $p$. In particular, $f(p)$ is a power of $p$ for all $p$ sufficiently large. Yet for sufficiently large inputs, $x^{d - 1} < f(x) < x^{d + 1}$ so that $f(p) = p^d$ for all $p$ large enough. Since $f$ is polynomial, it follows that $f(x) = x^d$ identically. It's easy to see that all these polynomials do in fact satisfy the problem condition.

Let $f(x)=a_0x^n+a_1x^{n-1} \cdots + a_n$. Suppose , $a_n >1$ . Then, there exist a prime factor of $a_n$ . Let it be $p$. Then $p|f(p^n) \forall n$. But suppose any of coefficient is $\neq 0$, then we can make $n$ sufficiently large, such that that term exceed, all terms with degree < than that term. Thus we get a contradiction. So all coeffecient $=0$. So the polynomial $f(x)=a_0x^n$ and by the property of this polynomial, we get $a_0=1$ Now if $a_n =1$ . Then we consider the polynomial $g(x) = f(f(x))$. Then notice that this polynomial also satisifies all the conditions of $f(x)$. And in this case the constant term $>1$. So proof follows from the case 1. Thus only polynomial is $f(x)=x^n$ And we are done. $\Box$

Apparently Romanian TST 2010 is famous.