Mongolian Mathematical Olympiad Round 3

Answer: There is no solution.
Our equation is equivalent to
$$(4z^2 + 1)^2 = (4x - 1)(2y - 1).$$
Since $4x - 1 \equiv 3 \pmod{4}$, there exists a prime divisor $p \ge 3$ of $4x - 1$ such that $p \equiv 3 \pmod{4}$. Then $p \mid 4z^2 + 1$, thus $4z^2 \equiv -1 \pmod{p}$. By Fermat's theorem, we have
$$1 \equiv (2z)^{p-1} \equiv (-1)^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$$
This is a contradiction. So there is no solution.