Solutions to diophantinte equation $x^4+2y^4=z^2$

Question

It is well known that the equation $x^4+y^4=z^2$ has no non-trivial solutions. The same holds also for the equation $x^4+2y^4=z^4$. What about the equation $x^4+2y^4=z^2$?

Answer 1

The classic book Diophantine Analysis by R. D. Carmichael says on page 17 that the equation $x^4+2y^4=z^2$ has no non-trivial solutions. Here is the original text, slightly edited:

4. The equation $x^4+2y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which are different from zero. Suggestion. This may be proved by the method of infinite descent. Begin by writing $z$ in the form $ z = x^2 + \frac{2py^2}{q}, $ where $p$ and $q$ are relatively prime integers, and thence show that $x^2=q^2-2p^2$, $y^2=2pq$, provided that $x$, $y$, $z$ are prime each to each.