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(x) $ is odd for left side of the equation and is even for right side of the equation.