The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$ Problem.  Let $a$ be a positive integer that is not a perfect cube. Show that the Diophantine equation $(xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers $x, y, z$ with $z > a^{2}+2a$.
If this result is true, we can prove that for a given integer $a$, $a \not= m^3$, there are finitely many Fermat pseudoprimes of the form $ap^{3}+1$ to any base $b>1$ where $p$ runs through the primes.
Theorem 1. Let $n = ap^{3}+1$, $a \not = m^3$, $p > a^2+2a$ is prime. If there exists an integer $b$ such that $b^{n-1} \equiv 1 \  ($mod $\ n)$ and $b^{a} \not\equiv 1 \  ($mod $\ n)$ then $n$ is prime.
Proof. It can be shown that if  $p \ | \ \phi(n)$ then $n=(sp+1)(tp+1)$ for some integer $sp+1$ and prime $tp+1$.
Because the Diophantine equation $(sp+1)(tp+1)=ap^3+1$ has no solutions in positive integers $s, t$ when $p > a^2 +2a$, we must have $s = 0$. Therefore if $p \ | \ \phi(n)$, $n = tp+1$ is prime.
Assume $n$ is composite, $b^{n-1} \equiv 1 \  ($mod $\ n)$ and $b^{a} \not\equiv 1 \  ($mod $\ n)$, we have $ord_n b \ | \ \phi(n)$. If $p \ | \ ord_n b$, then $p \ | \ \phi(n)$, a contradiction because $n$ is assumed composite therefore $ p \ \not | \ ord_n b$. We also have $ord_n b \ | \ n-1 = ap^3$. Because $ p \ \not | \ ord_n b$, we must have  $ ord_n b \ | \ a$. Hence $b^a \ \equiv 1 \ (mod \ n)$ which contradicts the hypothesis therefore $n$ must be prime.
Remark. For all $ n > b^a$, we have $b^a \ \not\equiv 1 \ (mod \ n)$ therefore there are finitely many Fermat pseudoprimes of the form $ap^3+1$ for a given positive integer $a$ that is not a perfect cube.
A lengthy and incomplete solution to this problem can be found here https://math.stackexchange.com/questions/3842292/prove-that-the-diophantine-equation-xz1yz1-az3-1-has-no-solutions-in
 A: The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $
Note that $z \ | \ x+y$, therefore $z \le x+y. \ $  Treating $x, y$ as constants, the only positive solution for $z \ $ is   \begin{equation} z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a}
\end{equation} In order for $z$ to be rational, the discriminant must be a perfect square. Therefore $w^2 = x^2y^2+4a(x+y)$. We see that $w > xy$ and $w \equiv xy \ ( $mod$ \ 2)$. We can write $w = xy + 2t$, $t > 0$.
Substituting $w$ above,  $(xy+2t)^2 = x^2y^2+4a(x+y)$. Expanding and simplifying, $txy - ax - ay +t^2 = 0$. Multiplying through by $t$ and factoring, $(tx-a)(ty-a)=a^2 - t^3$. We must have $t \le a-1$ otherwise $RHS<0$ and $LHS \ge 0$. Because $a$ is not a perfect cube, $a^2 - t^3 \not = 0$. The remainder of the proof utilizes the result: If $ab = c \  $ where $a,b, c \not = 0$ are integers then $a+b \le c+1$ if $c>0$ and $a+b \le -(c+1)$ if $c < 0$.
We now consider two cases:
Case $1: \ $ $ a^2 - t^3 >0 \ ;$
Using the result above on the factored equation, we have $(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$.
Hence, $z \le x+y \le (a^2+2a)/t \le a^2+2a \\\\$.
Case $2: \ $ $ a^2 - t^3 < 0 \ ;$
As in case $1$, we have $(tx-a)+(ty-a) \le t^3 - a^2-1 \ $,  $x+y \le t^2 - (a^2-2a+1)/t < t^2$.  Hence , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$
