Polynomial $P(x)$ with integer coefficients is given. For some positive integer $n$ numbers $P(0),P(1),\dots,P(2^n+1)$ are all divisible by $2^{2^n}$. Prove that values of $P(x)$ in all integer points are divisible by $2^{2^n}$.
If $P(i)\equiv 0\pmod{2^{2^n}}$ for $0\le i\le 2^n+1$ then $$P(x)\equiv x(x-1)\cdots (x-2^n-1)Q(x)\pmod{2^{2^n}}$$for some integral polynomial $Q(x)$.
Note that $(2^n+2)!\mid x(x-1)\cdots (x-2^n-1)$ since $\frac{x(x-1)\cdots (x-2^n-1)}{(2^n+2)!} = \binom{x}{2^n+2}\in \mathbb{Z}$. Meanwhile, the number of factors of $2$ that divide $(2^n+2)!$ is $$\nu_2((2^n+2)!) = \sum_{i\ge 1}\left\lfloor \frac{2^n+2}{2^i}\right\rfloor = 2^{n-1}+1 + \sum_{i=0}^{n-2} 2^i = 2^n.$$Thus, $x(x-1)\cdots (x-2^n-1)\equiv 0\pmod{2^{2^n}}$ identically, proving the claim.
I was PMed to add details, and I realized this problem does deserve some more careful explanation.
The first statement is true due to factor theorem:
$$P(a) \equiv 0\pmod{2^{2^n}} \iff P(x) = (x-a)Q(x).$$We can see this is true by applying long division of $x-a$ on $P(x) \equiv a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ and finding the remainder to be $P(a)\equiv 0$.
We apply factor theorem for $0\le i\le 2^n+1$ iteratively; we do need to ensure though that $x(x-1)\cdots (x-k+1)\not\equiv (x-k)Q(x)$ for any $0\le k\le 2^n+1$. We can use factor theorem once again to reduce this to confirming whether $k!\not\equiv 0$, which it isn't as $\nu_2(k!) < 2^n$ for all $k < 2^n+2$.