Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equation has infinitely many solutions in positive integers $x, y, z$ Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers..
From experiment, it appears the assertion is true for all polynomials $P(z)\equiv 1$(mod$ \ z)$ of degree $n>3$. How do we go about proving this?
Note if $n=3$, it has been shown that the Diophantine equation has a finite number of solutions
https://mathoverflow.net/questions/392002/is-xz1-a-proper-divisor-of-a-3z3a-2z2a-1z1-finitely-often/392018#392018

your first recipe, 5,20, 51, 104..  works
$$ z = n^3 + 2n^2 + 2n   $$
$$x = n+1 $$
$$ 1+xz = n^2 + 3 n^3 + 4 n^2 +2n+1 $$
$$ y = n^5 +3n^4 +5n^3 + 4 n^2 + 1 $$
$$  1 + yz = n^8 +5n^7 +13n^6 +20n^5 + 19n^4 + 10n^3 + 2 n^2 + 1$$
$$  ( 1+xz)(1+yz) = n^{12} + 8n^{11} + 32n^{10} + 81n^9 + 142n^8 + 178n^7 + 161n^6 + 104n^5 + 48n^4 + 17n^3 + 6n^2 + 2n + 1 $$
which is the same as $z^4 + z^3 + z^2 + z + 1.$
