Prove the equation $\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$ has no solution in the positive integers

Question

Prove the equation $$\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$$ has no solution in the positive integers

My work:

1) I have the usually problem

$$\left(nx^2+1\right)\left(my^2+1\right)=(m+n)z^2+1$$

in the positive integers. Initially I use case $\gcd(m,n)=1$

2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$

Answer 1

As it says in the comments this is answered on Math Overflow. The only solution is indeed the trivial one: $(0,0,0)$. This can be found in Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations.

Answer 2

The question is concerned only with $x,y,z\ge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $\mod x$.

$1\cdot (2y^2+1)\equiv 4z^2+1\mod x$.

Let $z\equiv a\mod x$. Then $2y^2+1=4(nx+a)^2+1$.

Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2\Rightarrow 2=\frac{y^2}{(nx+a)^2}\ $ which would make $\sqrt{2}$ rational. Hence there are no positive integers that satisfy the equation.