Here is the Theorem in my textbook:
For each nonconstant polynomial $f(x)\in\mathbb{Z}[x]$, the set of prime divisors of the integers $\{f(k):k\in\mathbb{N}_{0}\}$ is infinite, where $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ and $\mathbb{Z}[x]=\{a_{n}x^{n}+\dots+a_{0}:a_{j}\in\mathbb{Z}\}$.
Here's a proof as follows but I can't figure out the last line :
Suppose that $$f(x)=a_{0}+a_{1}x+\dots+a_{m}x^{m}$$ and assume that for the set $\{f(k):k\in{\mathbb N}_{0}\}$ the number of prime divisors that occur for some $f(k)$ is finite. Let $U=\{p_{1},\dots,p_{n}\}$ be this set of prime divisors and let $D=p_{1}\dotsm p_{n}$.
Without loss of generality, suppose $a_{0}\ne 0$. Choose an integer $t$ such that $p_{i}^{t}\nmid f(0)=a_{0}$ for all $i$. Since $p_{i}$ are the only primes we must have $a_{0}\mid D^{t}$, that is, $D^{t}=a_{0}b$ for some $b\in\mathbb{Z}$.
For $k\ge1$ we have $$ f(kD^{2t})=\sum_{j=1}^{m}a_{j}k^{j}D^{2tj}+a_{0} =a_{0}\biggl(\,\sum_{j=1}^{m}a_{j}k^{j}b^{2j}a_{0}^{2j-1}+1\biggr) =M $$
For $k$ large enough the integer $M$ must have a prime divisor $p$ that $p\nmid a_{0}b$ and hence $p\notin U$.
Any comment or advice I will be grateful.