Can I prove without using properties of the elliptic curves , that the equations $$x^2+y^3=z^6$$ $$x^2+y^6=z^3$$ $$x^3+y^6=z^2$$ have no solution in coprime positive integers except $(2,1,3)$ satisfying the third equation ?
How can I exploit somehow the fact that a $6$th power is also a perfect square and a perfect cube for an elementary proof ?
Motivation : This case completes the euclidean case in the generalization of Fermat's last theorem, where the exponents $p,q,r$ satisfy $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$. The case $(3,3,3)$ is covered by Fermat's last theorem and the case $(2,4,4)$ is well known to have no non-trivial solution.