Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?
To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.
Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.
A family of solutions is then given by $x=m$, $y=m^{2}z - m$  where $z$ takes on any positive integer.
How do I go about proving the striking observation: There are no positive integer solutions  $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?

Let $a$ be a positive integer that is not a cube, and let $x$, $y$ and $z$ be positive integers such that
$$(xz+1)(yz+1)=az^3+1.$$
Expanding the left hand side and rearranging a bit then shows that
$$az^2-xyz-(x+y)=0,\tag{1}$$
so $z$ is an integral root of a quadratic with discriminant $x^2y^2+4a(x+y)$. In particular this discriminant is a perfect square, so there exists a positive integer $v$ such that
$$x^2y^2+4a(x+y)=(xy+2v)^2,$$
and with a bit of rearranging we find the curious identity
$$(a-xv)(a-yv)=a^2-v^3.$$
We see that $v<a$ as otherwise the right hand side is negative, whereas the left hand side is not. Applying the quadratic formula to $(1)$ shows that
$$z=\frac{xy+\sqrt{x^2y^2+4a(x+y)}}{2a}=\frac{xy+(xy+2v)}{2a}=\frac{xy+v}{a},$$
where we have the $+$-sign because $z$ is positive. It follows that
$$z<\frac{xy}{a}+1,$$
so now to prove that $z<a^2+2a$ it suffices to show that $xy<a(a+1)^2$.
