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.
 A: 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
$$ a^9 t^{12} + 1  $$
=======================================
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
$$ x = y \left( y^2 z^2 - 1 \right)  $$
A: This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$

Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\
&=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right)
\end{align}$$
Where the last equation is because when $p$ is odd, $$
\begin{align}u^p+1&=(u+1)
\sum_{j=0}^{p-1} (-1)^ju^j
\\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$
Now, since $n>d,$ we can set $$
\begin{align}x&=a^{n-d}t^{k-p}\\
y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}
\end{align}$$
For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.
So this solves all cases $k>3.$

You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$
k even
So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$
Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$
k odd
Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$  Then for any integer $t$:
$$\begin{align}z&=a^{m-1}t^{2m-1}\\
x&=at^2\\
y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1}
\end{align}$$
is a solution.

In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$
