Let $S=\{p: p\mid f(n), n\in \mathbb{Z}_+, p \text{ is a prime}\}$, we know $|S|=+\infty$, so we can pick a large enough prime $p \in S$, $p > |f(0)|$, assume $p\mid f(k)$, but $p\nmid f(j), 1\leq j\leq k-1$. Now, $p\mid f(n+1)f(n+2)\cdots f(n+k)$ for any $n\in \mathbb{Z}_+$, we select $n\equiv -1 \pmod{p}$, then $p \mid f(0)f(1)\cdots f(k-1)$, it follows that $p \mid f(0)$ because $p\nmid f(j), 1\leq j\leq k-1$, now we get $f(0)=0$ because $p>|f(0)|$, which means $x\mid f(x)$.