I've got an equation of the form
$$ a^4+1=2b. \qquad(\star) $$
By well-known results regarding the sum of two squares, $b$ must be the sum of two squares. But does $(\star)$ force any other restrictions on $b$ as a result of the left-hand side being the sum of two fourth powers?
On
We get a little information about factors of $b-1$: $$2(b-1)=a^4-1=(a^2-1)(a^2+1)=(a+1)(a-1)(a^2+1)$$
On
Every prime factor $p$ of $b$ must be $\equiv 1 \pmod{8}$, since $a^4 \equiv -1 \pmod{p}$. For a sum of two squares, you only get that the odd prime factors of $b$ are $\equiv 1 \pmod{4}$ (or appear with an even power, but if one of the squares is $1$, that excludes any prime factor $\equiv 3 \pmod{4}$).
The number $1$ is the most restrictive thing. $b$ can only be divisible by primes such as $$ 17,41,73,89,97, \cdots $$ that is primes $p \equiv 1 \pmod 8.$
It is fairly quick to rule out primes $q \equiv 3 \pmod 4.$ I will see what I can do about a reference for primes for which $-1$ is a quartic residue satisfying $p \equiv 1 \pmod 8.$ EDIT: It is if and only if, and follows from the supplementary laws at the bottom of page 82, after Theorem 4.21 in Primes of the Form $x^2 + n y^2$ by David A. Cox.
The main thing is that there are far fewer numbers of your type up to some large bound $N,$ roughly $\sqrt {\sqrt N}.$ The count of sums of two squares up to $n$ is about $$ \frac{0.7642 \, N}{\sqrt {\log N}}, $$ which is a much larger figure.