When does $(xz+1)(yz+1)=az^{3}+1 $ have finitely many solutions in positive integers?

Question

Consider the diophantine equation in three variables $x$, $y$ and $z$; ($xz+1$)($yz+1$) $=$ $6z^{3}+1$. The only positive integer solutions I have found are {$x=4,y=10,z=7$} and {$x=10,y=4,z=7$}. From a Maple program, I have iterated over all values of $z$ in the range $50<z<10^{8}$, the only corresponding solutions of $x$ and $y$ are those with $ x=0$ and $y$ positive and vise versa. I would like to find out if this diophantine equation contains finitely many or infinitely many solutions in positive integers $x, y$ and $z$. In general; For a given positive integer $a$, what conditions are sufficient for the diophantine equation ($xz+1$)($yz+1$) $=$ $az^{3}+1$ to have finitely many solutions in positive integers $x, y$ and $z$. From experimental results, it appears that this equation has finitely many solutions in positive integers if and only if $a$ is not a third power of any integer i.e. $a\neq m^{3} $ for all integers $m$. Any help or references on this question will be appreciated.

Answer 1

Here is a partial answer: If $a=b^3$ is a cube, then there is an infinite family of solutions to $(xz+1)(yz+1)=az^3+1=b^3z^3+1$ given by $$(x,y,z) = (b, b^2z-b, z),\ b, z\in\mathbb{N}.$$ This arises from the factorization $b^3z^3+1 = (bz+1)(b^2z^2-bz+1) = (bz+1)((b^2z-b)z+1)$.

In addition to the above, for any $a$ there are solutions $(a+1, a^2+a-1, a^2+2a)$ and $(2a-1, 2a+1, 4a)$, and there appear to be (empirically) solutions for some values of $x$ between $a+1$ and $2a-1$. For all of these solutions, $z = x+y$ and each $x$ corresponds to a unique $y$. There appear to be no solutions for $y\ge x>2a-1$. These, together with a finite set of solutions for $x<a+1$, appear to cover all solutions to the equation. I can prove very little of this.