I tried the heuristic from the Hardy-Littlewood circle method for this equation. The heuristic suggests that the number of solutions within the range $\max\{\vert x\vert,\vert y\vert,\vert z\vert\} should be something like
$\sigma_{\infty}\prod_{p}\sigma_p,$
where $\sigma_{\infty}$ and $\sigma_{p}$ are real density and local densities.
$\sigma_{\infty}=\lim_{\epsilon\rightarrow 0}\frac{1}{2\epsilon}\vert\{(x,y,z)\mid\max\{\vert x\vert,\vert y\vert,\vert z\vert\}
$\sigma_p=\lim_{n\rightarrow \infty}\vert\{(x,y,z)\mid x^4+y^4+1\equiv z^2\mod p^n\}\vert/p^{2n}$
These densities have explicit formulas(if I calculated them correctly)
if $p\equiv 3 \mod 4$, $\sigma_p=1-1/p.$
if $p\equiv 1 \mod 4$, $\sigma_p=1+\frac{1+6(-1)^{(p-1)/4}}{p}-\frac{2a}{p^2},$
where $p=a^2+b^2$, $a\equiv 3\mod 4$.
$\sigma_2=1$ and $\sigma_{\infty}\approx\frac{B(1/4,1/4)}{2}\log 2N\approx 3.70815\log 2N,$ where $B$ is Euler beta function.
The infinite product over prime numbers $p$ is not absolutely convergent, but it is indeed convergent.
$\prod_{p}\sigma_p\approx 0.0193327$.
$(\pm x,\pm y,\pm z)$ and $(\pm y,\pm x,\pm z)$ are all solutions of the Diophantine equation, so the number of essentially different solutions should be $\frac{B(1/4,1/4)}{32}\prod_{p}\sigma_p\cdot\log 2N\approx 0.0044805\log 2N.$ So, if this heuristic works, the first solution may occur near $N\sim\exp(1/0.0044805)/2\approx 4\times 10^{96}$, which means $x,y\sim 10^{48}$. It is possible that the solutions are enormous, but I am not sure if there are other obstructions to the equation.
Edit: I guess things are different for Euler's quartic $x^4+y^4+z^4=w^4.$ The real density for this surface is still about $c\log N$, but the infinite product over primes diverges to $\infty$. The infinite product grows like $(\log N)^r$, where $r$ is some real number. So I think the integral points on Euler's quartic is quite "dense" compared to $x^4+y^4+1=z^2$.