Prime divisors of sequences of integers I have the following problem and I need your support for it

Problem. Let P(x) be a polynomial with integer coefficients, such that $\deg P>0$ and $\lim_{x\to+\infty}P(x)=+\infty$. Prove that there exist infinitely many prime numbers $p$ such that for some natural number $n$
  $$p\mid \left\lfloor \log\left(2017^{P(n)}+1\right)\right\rfloor.$$

 A: As I pointed out in my comments, what matters here is the growth. First we prove :

Proposition : Given $f : \mathbb{N} \mapsto \mathbb{N}$ strictly increasing, and $k \ge 0$ such that $f(n) = O_{n \to +\infty} \big( n^k \big)$, the set $\{ p \mbox{ prime } | \ \exists n:\ p|f(n) \}$ is infinite.

Proof : ad absurdum, assume we can write this set $p_1, ..., p_r$. Let us denote $A = \{p_1^{k_1} \cdot\cdot\cdot p_r^{k_r}\ |\ k_i \ge 0\}$, $B = f(\mathbb{N})$, and for all $n$, $A_n = A \cap [\![1, n]\!]$ and $B_n = B \cap [\![1,n]\!]$. As $B_n \subset A_n$, $|B_n| \le |A_n|$. 
However, as all $p_i$ are $\ge 2$, $|A_n| \le (\mbox{log}_2(n)+1)^r$.
And as $f$ is strictly increasing, $|B_n| = \max(\{m | f(m)\le n\})$. It is easy to conclude that $n^{1/k} = O_{n \to +\infty} (|B_n|)$.
Conclusion : $n^{1/k} = O_{n \to +\infty}\big( (\mbox{log}_2(n)+1)^r\big)$, which is absurd.

Regarding your problem, it is easy to check that $f :n \mapsto \left \lfloor \mbox{log}\Big( 2017^{P(n)}+1 \Big)\right \rfloor$ is strictly increasing (the quantity between the floor brackets increases by more than $\mbox{log}(2017)>1$ between $n$ and $n+1$).
Moreover, $f(n)\le P(n)*\mbox{log}(2017)+\mbox{log}(2017)+1$. We can apply the proposition.
