Let $m$ be a positive integer such that $p:=8m^2+1$ is prime.
Conjecture : We always have $$2^{2m^2}\equiv 1\pmod p$$
I could only establish $$2^{4m^2}\equiv 1\pmod p$$ following from Euler's criterion , but I could not rule out $2^{2m^2}\equiv -1\pmod p$
Motivation : A proof would show that $$(2m^2+1)(8m^2+1)$$ is a Poulet-number whenever both factors are prime.
Verification : The conjecture is true upto $m=10^{10}$.
The conjecture follows if we can prove that $p$ splits totally in $K=\mathbb{Q}(i, \sqrt[4]{2})$: indeed, there would be a ring homomorphism $\mathcal{O}_K \rightarrow \mathbb{F}_p$; it would map $2=(\sqrt[4]{2})^4$ to a fourth power!
Note that $p \equiv 1 \pmod{8}$, so we already know that $p$ splits in the slightly smaller field $\mathbb{Q}(i,\sqrt{2})$.
Let $\mathcal{O}=\mathbb{Z}+2R$ be the order of $R=\mathbb{Z}[\sqrt{-2}]$ of conductor $2$.
The hypothesis says that $p$ is the norm of an element $u \in \mathcal{O}$. It follows by class field theory that $p$ splits in the ray class field $H$ of modulus $2$ of $C=\mathbb{Q}(\sqrt{-2})$.
Because $R$ is a PID, it’s elementary to show that $H/C$ has degree $4$.
Because $K/C$ is only ramified above $2$, it’s easy to check (by working only at $2$) that the norm of the idèle class group of $H$ is contained in the norms of the idèle class groups of $C(i)$ and $C(\sqrt[4]{-2})$.
In other words, $H$ contains $C(i)$ and $C(\sqrt[4]{2})$ so $H$ contains $K$. By counting degrees over $C$, we find $H=K$, QED.