Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation
$(xz+1)(yz+1)=az^{k}+1$
has infinitely many solutions in positive integers $x,y, z$.
To motivate the question, it can easily be shown that if $k <3$, the given diophantine equation has no solutions in positive integers $x, y ,z$ with $z>a$.
Proof: $(xz+1)(yz+1)=az^{k}+1$ may be simplified to $xyz^{2}+(x+y)z=az^{k}$. If $k=1$, this reduces to $xyz+x+y=a$. Its clear that $a>z$ therefore there are no positive integral solutions in $x$ and $y$ when $z>a$. if $k=2$, we have the reduced equation $xyz+x+y=az$. We have $z$ | $x+y$, $z \le(x+y) \le xy$. Therefore $LHS=xyz+x+y>z^{2}$. Because $RHS=az$, we must have $a>z$ thus there are no solutions in positive integers $x ,y$ when $z>a$.
I would like to prove that given two positive integers $a$ and $k>3 $, the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many positive integer solutions $x, y, z$. I do not know how to start the proof.