Actually you can prove a lot more.

**Theorem**

For any non-constant polynomial $p(x)\in\mathbb{Z}[x]$ and any positive integer $k$ there is an integer $n$ such that $p(n)$ is divisible by at least $k$ distinct primes.

**Proof**

If we prove that there exist integers $n_1,\ldots,n_k$ and distinct primes $p_1,\ldots,p_k$ such that $p(n_i)\equiv 0 \bmod{p_i}$ then we are done, because there exists an $n$ such that $n\equiv n_i\bmod{p_i}$ by the Chinese Remainder Theorem, and any such $n$ satisfies $f(n)\equiv f(n_i)\equiv 0 \bmod{p_i}$, as desired.

Now by contradiction suppose that $p$ is only divisible by the primes $p_1,\ldots,p_l$, $l\le k-1$. Since we have that $p(0)\neq 0$, let $p(0)=\pm p_1^{\alpha_1}\ldots p_l^{\alpha_l}$, $x\equiv 0\bmod{p_1^{\alpha_1+1}\ldots p_l^{\alpha_l+1}}$.

Then $p(x)\equiv p(0)\bmod{p_i^{\alpha_i+1}}$, $1\leq l\leq k-1$, so that the greatest power of $p_i$ that divides $p(x)$ is $p_i^{\alpha_i}$.

But by hypothesis $p(x)$ is only divisible by the $p_i$, so we conclude that $p(x)=\pm p(0)$. Using the pigeonhole principle and the fact that a non-constant polynomial can only assume a value a finite amount of times this is a contradiction, as desired.