The exponential generating function of the unsigned Euler numbers $E_{n}$ is $$\frac{1}{\cos x}=\sum_{n\ge 0}\frac{ E_n}{n!}x^n$$ For $k,n\gt0$, not too large, one can observe that $$E_{ 2^{2^k}\cdot n} \equiv 0 \mod (2^{2^k}+1) $$ For $k,n\gt0$, I would say more:
$$E_{ 2n} \equiv 0 \mod (2^{2^k}+1) \Longleftrightarrow n \equiv 0 \mod2^{2^k} $$
Is this true?
UPDATE: this is true in the $\Longleftarrow$ direction, when $2^{2^k}+1$ is prime since it can be shown that
$$E_{2 m}\equiv 0 \ \ \bmod \prod_{\begin{array}{c} p\equiv 1 \bmod 4 \\ p-1\mid 2m \end{array}}p$$
where $p$ is a prime and the product is $1$ when there is no factor.
But this is wrong in general because it can also be shown that
$$E_{2 m+p-1}\equiv E_{2 m} \ \ \bmod p $$
where $p$ is odd prime.
$2^{2^5}+1=641\cdot 6700417$
$2^{2^5} \equiv 256 \ \ \bmod 640 $
then $E_{2^{2^5}}\equiv E_{256}\equiv 560 \ \ \bmod 641 $
therefore $E_{2^{2^5}}$ is not divisible by $641$, hence $E_{2^{2^5}}$ is not divisible by $2^{2^5}+1$ either.