<s>[INST] Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers. 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.
 [/INST] Getting there. Here is $k=4.$   a family of solutions to
$$  a z^4 + 1 = (xz+1)(yz+1)  $$
is parametrized by integer $t$ with
$$  y=at $$
$$ x = a^4 t^5 - at  $$
$$ z = a^2 t^3  $$
Both sides of the equation are
$$ a^9 t^{12} + 1  $$
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For that matter, we can take care of all $k \neq 0 \pmod 3$  this way.
When $k > 3$  and $k \equiv 1 \pmod 3,$  we may take
$$  y = a^{\frac{2k-5}{3}} \; t^{k-3}  $$
$$ z = a^2 t^3  $$
followed by
$$ x = y \left( y^2 z^2 - 1 \right)  $$
When $k > 3$  and $k \equiv 2 \pmod 3,$  we may take
$$  y = a^{\frac{k+1}{3}} \; t^{k-3}  $$
$$ z = a t^3  $$
followed by
$$ x = y \left( y^2 z^2 - 1 \right)  $$
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