The solutions of the diophantine equation $z^x=1+(z-1)y^2$

Question

I'm trying to determine the positive integer solutions of the diophantine equation $z^x=1+(z-1)y^2$. This equation can be written also as $$y^2=\frac{z^x-1}{z-1}=\sum_{i=0}^{x-1}z^i$$ Trivial solutions $(x,y,z)$ are: $$(0,0,m)\;\;\;(1,1,m)\;\;\;(2,m,m^2-1)$$ Other two solutions are: $$(5,11,3)\;\;\;(4,20,7)$$ Is it possible to prove that all the solutions are the ones mentioned above?