Cross-post: This very elementary question was first posted to Mathematics Stack Exchange but the response I got there (even after offering a bounty) was not useful.

For the purpose of this question, a *pseudoprime* is a composite number $n$ satisfying $2^{n-1} \equiv 1 \ (\text{mod}\,n)$, also known as a (composite) odd, weak Fermat pseudoprime to base two.

If we look at $n$ of the form $n=b^2+1$ we find many primes ($b$ in OEIS A005574) and pseudoprimes (A135590).

However, if we move to $n=b^4+1$ we still find many primes (A000068), but the only pseudoprimes we have been able to find are of the form $b=2^{2^k}$ which makes $n=b^4+1$ a Fermat number (which is obviously either prime or pseudoprime).

Question:If $b^4+1$ is composite but we still have $2^{b^4} \equiv 1 \ (\text{mod}\ b^4+1)$, will $b$ necessarily be of the form $2^{2^k}$?

If there is no obvious reason why this should be true, can someone provide some heuristics on the "expected" asymptotic behavior of such numbers $b$? A computer search seems to demonstrate that there are no examples $b \le 2\cdot 10^{10}$. Maybe there is a smarter way to locate an example?

Or maybe this has been asked/answered before in the literature?