Finding solutions to $(4x^2+1)(4y^2+1) = (4z^2+1)$ Consider the following equation with integral, nonzero $x,y,z$
$$(4x^2+1)(4y^2+1) = (4z^2+1)$$
What are some general strategies to find solutions to this Diophantine?
If it helps, this can also be rewritten as $z^2 = x^2(4y^2+1) + y^2$
I've already looked at On the equation $(a^2+1)(b^2+1)=c^2+1$

Let $a$ be a positive integer.
Then
\begin{align}
(4a^2+1)(4((2a)^2)^2+1) &= 256a^6 + 64a^2 + 4a^2 + 1  \\
& = 4(64a^6 + 16a^4 + a^2) + 1 \\
&= 4(a^2(8a^2+1)^2)+1  \\
&= 4((8a^2+1)a)^2+1
\end{align}
so $(a, (2a)^2, (8a^2+1)a)$ is always a solution.
There are others as well.
