Are there any rational numbers $x, y, z$ with $xyz \neq 0$ such that $y^2 = x^3 + z^{4k}$ for some $k \in \mathbb{Z}\_{>1}$ ?



On the elliptic curve $y^2 = x^3 + z^{4k}$
For $k=1$, the surface $y^2=x^3+z^4$ is a rational surface, so it has lots of rational points. The substitution $x=zu$ and $y=z^2v$ leads to $z=u^3/(v^2-1)$, so for almost all $u,v\in\mathbb Q$, the point
$$
\left( \frac{u^4}{v^2-1},\; \frac{u^6v}{(v^2-1)^2},\; \frac{u^3}{v^2-1} \right)
$$
satisfies $y^2=x^3+z^4$. (The same thing should work in general, take $x=z^eu$ and $y=z^{2k}v$, where $3e=4k\pm1$, with the sign chosen so that $e$ is an integer.)


