Similar to finishing
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
where I had an acceptable bound but needed help from Gerry Myerson to improve to the sharp bound.
We have $$ (xz+1)(yz+1) = a z^3 + 1 $$
This becomes
$$ a z^3 - xyz^2 - (x+y)z=0$$
or
$$ a z^2 - xyz - (x+y) = 0 $$
We get
$$ z = \frac{ xy + \sqrt{ x^2 y^2 + 4a(x+y) } }{2a} $$
Let me also record
$$\color{fuchsia}{ z(az-xy) = x+y }$$
which follows directly from $ a z^2 - xyz - (x+y) = 0 $
Note also the simple
$$\color{fuchsia}{ z \leq x+y }$$
It is necessary to have square discriminant to get a rational value for $z,$ take
$$ w^2 = x^2 y^2 + 4a(x+y) $$
We have $$ w > xy $$ and
$$ w \equiv xy \pmod 2. $$
Therefore we can define an integer $t,$ when it all works, with
$$ w = xy+2t $$
Now $$ z = \frac{xy+w}{2a} = \frac{xy+xy+2t}{2a} = \frac{2xy+2t}{2a} = \frac{xy+t}{a} $$
$$ z = \frac{xy+t}{a} $$
There are always three flavors for any $a$
$$ t=a-1 \; , \; y = 1 \; , \; x = a^2 - 3a +1 \; , \; z = a-2 $$
$$ t=1 \; , \; y = 2a-1 \; , \; x = 2a +1 \; , \; z = 4a $$
$$ t=1 \; , \; y = a+1 \; , \; x = a^2 +a -1 \; , \; z = a^2+2a $$
From
$$ x^2 y^2 +4a(x+y) = (xy+2t)^2 $$
we get $$ t xy - ax -ay + t^2 = 0, $$
$$ t^2 xy - tax -tay + t^3 = 0, $$
$$ \color{red}{(tx-a)(ty-a) = a^2 - t^3} $$
IF $a > 1$ and $t = a + \delta$ with $\delta \geq 0,$ we find
$$ ((a+\delta)x-a)((a+\delta)y-a) = a^2 - (a+\delta)^3 < 0 $$
since $a>1.$
However, the left hand side is non-negative, which is a contradiction.
$$ \color{red}{ t \leq a-1} $$
I will fill in the (lengthy) details in a bit.
I always have $x \geq y \geq 1$
IF $$ \color{blue}{ a^{2/3} < t \leq a-1} $$
we get
$$ (tx-a) (a-ty) = t^3 -a^2 > 0 $$
so $a-ty >0,$ $ty - a < 0,$
$$ ty < a $$
$$ y < \frac{a}{t} < a^{1/3} $$
$$ a - ty \geq 1 $$
$$ tx-a \leq t^3 - a^2 $$
$$ tx \leq t^3 - (a^2 - a)$$
$$ x \leq t^2 - \frac{a^2 - a}{t} $$
DETAIL: As $t$ increases, $t^2$ increases, while $\frac{1}{t}$ decreases. Then $\frac{-1}{t}$ increases. We have $a \geq 2$ so that $a^2 - a > 0,$ so that
$\frac{-(a^2-a)}{t}$ increases. Together, $t^2 - \frac{a^2 - a}{t}$ increases and takes its maximum value at $t=a-1,$ that being $ a^2 - 3a + 1.$
Thus
$$ \color{magenta}{x \leq a^2 - 3a + 1}$$
$$xy + t < a^{7/3} -3a^{4/3} + a + a^{1/3} -1 $$
$$ z < a^{4/3} -3a^{1/3} + 1 + a^{-2/3} -\frac{1}{a} $$
$$ \color{red}{ z < a^{4/3} } $$ when $a^{2/3} < t \leq a-1$
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I always have $x \geq y \geq 1$
IF $$ \color{blue}{1 \leq t < a^{2/3} }$$
$$ (tx-a) (ty-a) = a^2 - t^3 > 0$$
$$ (tx-a) \leq a^2 -t^3$$
$$ tx \leq a^2 + a - t^3 < a^2 + a$$
$$ x \leq \frac{a^2 + a}{t} $$
Meanwhile
$$ t^2 xy - ta(x+y)= -t^3 < 0 $$
$$txy < ta(x+y) \leq 2ax $$
$$ ty < 2a
$$ y < \frac{2a}{t} $$
Together
$$ xy < \frac{2 a^3 + 2a^2}{t^2} $$
$$ xy + t < \frac{2 a^3 + 2a^2}{t^2} + a^{2/3} $$
$$ z < \frac{2 a^2 + 2a}{t^2} + \frac{1}{a^{1/3}} $$
IF $t \geq 2$
then $z < \frac{a^2 + a}{2}$
IF $t=1$ we have $$ (x-a)(y-a) = a^2 - 1 > 0 $$
If $x>a$ then $y>a.$ Then $y-a \geq 1$ and $x-a \leq a^2 - 1$
When $t=1$ we have $x \leq a^2 + a - 1.$
In general, when we have real $p \geq 1, q \geq 1,$ and $pq=c,$
the maximum of $p+q$ occurs when $p=1$ and $q=c$
so that $p+q \leq 1+c$
With $ (x-a)(y-a) = a^2 - 1 $ we get $x-a+y-a \leq a^2.$ Thus
$$ x+y \leq a^2 + 2a$$
With $t=1,$ we know $z = x+y.$ With $t=1$
$$ \color{red}{z \leq a^2 + 2a } $$
DETAIL $$\color{fuchsia}{ z(az-xy) = x+y }$$
and $$ z = \frac{xy+t}{a} $$
so that when $t=1,$ we get $az = xy+1$ or $az-xy = 1,$ so that $z(az-xy) = x+y$ tells us $z=x+y,$ when $t=1$
Claim
$ax + by = c$ has infinite number of positive solution if and only if $a$ and $b$ are of opposite signs and $\gcd(a,b)|c$
Proof
WLOG let $a>0$ and $b<0$. There exists a solution because $\gcd(a,b)|c$.
If $(x',y')$ is a solution then $(x'-b,y'+a)$ is a solution so we get a increasing ordered pair of $(x,y)$ that satisfies the equation. Hence there are infinite number of positive solutions for this case.
It is not possible if $a$ and $b$ are of same sign. Suppose there are infinitely many solution when $a$ and $b$ are of same sign then there exists a solution $(x',y')$ with $\mid x'\mid>c$ and $\mid y'\mid>c$ , but this implies $\mid ax'+by'\mid>c$ thus leading to a contradiction.
For example $(a,b,c)=(rt,-rt,2rt)$ where $r$ is a constant and $t$ is a integer variable will give you different values for $a,b,c$ such that the equation has infinitely many solutions.
Best Answer
This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\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 $x\mid z.$