For each positive integer $n$ such that $n! + 1$ is a power of $P(n)$, we write $n! + 1 = P(n)^{k_n}$ and in this case call $n$ good. Due to the left-hand side and $P(x) \ge x$ for $x$ big enough, we have $k_n < n$ and $P(n)$ must be odd for $n \ge 2$. Hence, using the LTE, we obtain
$$\nu_2 (P(n)^{k_n} - 1) \le \nu_2(P(n) - 1) + \nu_2(P(n) + 1) + \nu_2(k_n).$$
Hence at least one of $\{\nu_2(P(n) - 1), \nu_2(P(n) + 1), \nu_2(k_n)\}$ is at least $\frac{1}{3}\nu_2(n!)$ for good $n$. As there are infinitely many $n$, by pigeonhole principle, we now know that one of the following assertions is true:
* There are infinitely many $n$ such that $\nu_2(P(n) - 1) \ge \frac{1}{3}\nu_2(n!)$.
* There are infinitely many $n$ such that $\nu_2(P(n) + 1) \ge \frac{1}{3}\nu_2(n!)$.
* There are infinitely many $n$ such that $\nu_2(k_n) \ge \frac{1}{3}\nu_2(n!)$.
We also know that $\nu_2(n!) \ge \frac{n}{2}$. So if the first or the second assertion is true we know that for infinitely many $n$ we must have $P(n) > 2^{\frac{n}{6}}$ which contradicts the polynomial asymptotic behavior. If the last assertion is true then we also have for sufficiently large $n$, we have $k_n > 2^{\frac{n}{6}} > n$. $\square$