Positive integers such that $(x+y)(xy-1)=z^2+1$

Question

Do there exist positive integers $x,y,z$ such that $$ (x+y)(xy-1)=z^2+1 $$

In my previous question Can you solve the listed smallest open Diophantine equations?, I discuss the smallest equations for which we do not know if they have any integer solutions. Some equations (think about Fermat Last Theorem) have some trivial integer solutions, and the correct question to ask is whether they have a solution in positive integers. The equation in the title is the smallest one for which I do not know the answer. It is known that all positive divisors of $z^2+1$ must be $1$ or $2$ modulo $4$, but if we take $x$ and $y$ to be $2$ and $3$ modulo $4$, this condition is satisfied, so no contradiction. On the other hand, search returns no results.

Answer 1

As you have already noticed, we may assume that $x \equiv 3 \pmod{4}$, $y \equiv 2 \pmod{4}$. Let $p \equiv 3 \pmod{4}$ be a prime divisor of $y + 1 \equiv 3 \pmod{4}$ such that $\nu_p(y+1) \equiv 1 \pmod{2}$. Consider the equation $$(x + y)(xy - 1) = z^2 + 1$$ modulo $p$. $$(x + y)(xy - 1) \equiv (x - 1)(-x - 1) \equiv -x^2 +1 \pmod{p}$$ $$x^2 + z^2 \equiv 0 \pmod{p}$$ Therefore $p \mid x, z$ since $-1$ is a quadratic nonresidue modulo $p$. Let $Y = y + 1$. $$x^2Y - x^2 + xY^2 - 2xY - Y = z^2$$ $$Y(x^2 + xY - 2x - 1) = x^2 + z^2$$ There is a contradiction, because $\nu_p(\cdot) $ is odd for left side of the equation and is even for right side of the equation.