$f(x)=nx^m$

Lemma For each $f\in\mathbb{Z}[x]$ let $A(f)$ be the set of prime divisors of $f(x)$ when $x$ runs over the integers. If $f$ has positive degree then $A(f)$ is infinite. Proof of lemmaThe result is clear for all $f$ with $f(0)=0$. Suppose for the sake of contradiction that $A(f)$ is finite for some $f$ with nonzero independent term. Let $P\subseteq A(f)$ be the set of primes which divide $f(x)$ for any integer $x$ and let $Q=A\setminus P$. Let $b$ and $c$ be integers such that $bc=f(0)$, $(b,c)=1$, and $p|b$ for any $p\in P$. For each $q\in Q$ there exists an integer $r_q$ such that $q\not| f(r_q)$, because of the chinese remainder theorem there exists an integer $d$ with $b^2d\equiv r_q\mod q$ for each $q\in Q$. Now $f(x)=bc+xg(x)$ with $g(x)\in\mathbb{Z}[x]$ and $f(b^2d)=b(c+bdg(b^2d))$. Since $f(b^2d)\equiv r_q\mod q$ we have that $q\not| f(b^2d)$ for any $q\in Q$. Also $(p,c+bdg(b^2d))=1$ for any $p\in P$, so $c+bdg(b^2d)$ does not have prime factors in $A(f)$. As there exist inifinitely many such integers $d$ it is sure that for some $d$, $c+bdg(b^2d)\not=\pm 1$ so it does have prime factors outside $A(f)$ and this is the contradiction. (Returning to the problem...) Let $f$ be one such polynomial ($f(p)$ has at most $k$ disctintc prime divisors) and rewrite it as $f(x)=x^mg(x)$, with $g\in\mathbb{Z}[x]$ and $g(0)\not= 0$. Suppose for the sake of contradiction that $g$ has positive degree. We can apply the lemma to $g$ so $A(g)$ is inifinite. In particular we can find $k+1$ primes in $A(g)$ not dividing $g(0)$ and let $s$ be their product. For each of these primes $q$ there exist and integer $r_q\not\equiv 0\mod q$ such that $q|f(r_q)$. By the chinese remainder theorem there exists $a\mod s$ such that $a\equiv r_q\mod q$ for each of these $q$. Since $r_q\not= 0\mod q$ we have $(a, s)=1$ then Dirichlet's theorem on primes in arithmetic progressions asures the existence of infinitely many primes $p\equiv a\mod s$. Choose one of these $p$ such that $f(p)\not= 0$ and we have that $s|f(p)$ so $f(p)$ has more than $k$ prime divisors, this is a contradiction. So the only polynomials $f\in\mathbb{Z}[x]$ with the mentioned property are $f(x)=nx^m$