**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