I do have an algorithm that keeps tings in reasonable size. I'm going to use other letters.
You are looking at
$$ (u^2 + 1)(v^2 + 1)= w^2 + 1 $$
Fix any integer $u \geq 1.$ Next check all $-1-u \leq v \leq 1+u$
for solutions to
$$ w^2 - (1+u^2) v^2 = u^2 $$
From what I can see, there are at most 11 such $(w,v)$ pairs.
Next, the relevant automorphism matrix is
$$
\left(
\begin{array}{cc}
2 u^2 + 1 & 2 u^3 + 2u \\
2u & 2 u^2 + 1 \\
\end{array}
\right)
$$
That is, for each such $(w,v)$ pair, you get a new solution from
$$ (w,v) \mapsto \; \; \; \color{red}{ \left( \; \; \; (2u^2+1) w + ( 2u^3 + 2u) v \; , \; \; 2uw + ( 2 u^2 + 1) v \; \; \; \right)} $$
The reason to begin with some negative $v$ is to catch surprises. The predictable small positive solutions are $(w=u, v=0)$ and $(w = u^2-u+1, v=u-1)$
and $(w = u^2+u+1, v=u+1)$ there are occasional surprises, beginning with $u=8, 12, 18, 21,...$ Let's see, for a fixed $u,$ you take $|v| \leq 1+u,$ check for when $u^2 + (u^2 + 1)v^2$ is another square, call that $w^2.$ Put pairs $(w,v)$ and $(w,-v)$ into a list ordered by $v$
u=203
w^2 - 41210 v^2 = 41209 = 7^2 29^2
Thu Aug 27 18:28:05 PDT 2020
w: 203 v: 0 SEED KEEP +-
w: 837 v: 4 SEED KEEP +-
w: 41007 v: 202 SEED KEEP +-
w: 41413 v: 204 SEED BACK ONE STEP 41007 , -202
w: 2059663 v: 10146 SEED BACK ONE STEP 837 , -4
w: 16731057 v: 82418 SEED BACK ONE STEP 203 , 0
back step : 3 u: 203
Thu Aug 27 18:28:06 PDT 2020
w^2 - 41210 v^2 = 41209 = 7^2 29^2
Alright, by Cayley-Hamilton, the solutions split up into a small number of orbits of Fibonacci type,
$$ v_{j+2} = (4u^2 + 2)v_{j+1} - v_j $$
When $u=8,$ we have
$$ v_{j+2} = 258 \; v_{j+1} - v_j $$
$$ -128, 0, 128, 33024, 8520064,.. $$
$$ -30, 2, 546, 140866,.. $$
$$ -9, 7, 1815, 468263... $$
$$ -7, 9, 2329, 600873, ... $$
$$ -2, 30, 7742, 1997406,... $$
$$ 0, 128, 33024, 8520064,.. $$
$$ 2, 546, 140866, 36342882,...$$
$$ 7, 1815, 468263, 120810039,...$$
$$ 9, 2329, 600873, 155022905,... $$
As you can see, there is considerable repetition, and the $v$ values grow rapidly in each sequence. Writing it this way, the fixed $u$ value and a $v$ value give
$w = \sqrt{u^2 + (u^2 + 1)v^2}$
You will find that sticking to one of the $v$ sequences causes $w$ to obey the same rule,
$$ w_{j+2} = (4u^2 + 2)w_{j+1} - w_j $$
About when to stop, as $v$ gets large we find $w \approx uv.$
BUT WAIT, THERE'S MORE.
The predictable small non-negative solutions to $w = \sqrt{u^2 + (u^2 + 1)v^2}$ are $w=u, v=0,$
then $w= u^2 - u+1, v = u-1$ and $w = u^2 + u + 1, v = u + 1.$ One of the surprise additions comes when $u = 2 t^2,$ in which case $w = 2t^3 + t , v = t$
