Answer: The statement is true. Proof: For sake of contradiction, we assume the statement was false. Then there are only finitely many primes $p_1,p_2,...,p_m$ that divide $P(n)$, which we define as $a_zx^z+a_{(z-1)}x^{(z-1)}+...+a_1x+a_0$ with (possibly negative) integers $a_i$ (so $P$ is of degree $z$), for positive integers $n$. Let $p_1,...,p_k$ be the primes of these that do not divide $a_0$ (it is possible that $k=0$) and let $q$ be the product of these primes (possibly $1$). We now look at $P((|a_0|^a)*q)$ for arbritary posite integers $a \ge 2$. This number then is divisible exactly once by $a_0$ (as every other term in the sum is divisible by $(a_0)^2$ at least) and for the same reasoning exactly as often as $a_0$ divisible by the primes $p_{k+1},...,p_m$ that divide $a_0$. Since $q$ does not divide $a_0$, but every other term in the sum is divisible by $q$, $P((|a_0|^a) \cdot q)$ is not divisible by any factor of $q$, that is, none of $p_1,...,p_k$. Now our assumption is that $P((|a_0|^a) \cdot q)$ is not divisible by any prime not on our list (as $(|a_0|^a) \cdot q$ is a positive integer); so as of the above reasoning, it must be divisible exactly by $a_0$ and no other number, that is, $P((|a_0|^a) \cdot q)=\pm a_0$, which means that $|a_0|^a \cdot q$ is either a root of $Q$ where $Q(n)=P(n)-a_0$ or a root of $R$ where $R(n)=P(n)+a_0$. But as $z$ is finite (otherwise, $P$ would not be a polynomial), there are only finitely many roots to $Q$ and $R$ as they both have degree $z\ge 1$ and every polynomial of degree $z$ at least $1$ has at most $z$ integer roots. On the other hand, as a was an arbitrary integer greater than or equal to $2$, we can find infinitely many terms of the form $(|a_0|^a) \cdot q$, precisely, at least $2z+1$. But these can not all be roots of either of the polynomials $Q$ and $R$, as both of them can have at most $z$ roots, but $2z+1>z+z$. So we achieved our contradiction.