The surface $ (x^{2} + y^{2} - 2 y + 1) \cdot (x^{2} + y^{2} + 2 y + 1) \cdot (x^{2} + y^{2} - 2 x + 1) \cdot (x^{2} + y^{2} + 2 x + 1) - z^2 = 0$
has the points $\left(\dfrac{t}{s},\dfrac{t}{s},\dfrac{s^{4} + 4 t^{4}}{s^{4}}\right)$, $ \left(\dfrac{- s^{2} + t^{2}}{s^{2} + t^{2}},\dfrac{2 s t}{s^{2} + t^{2}} ,\dfrac{8 s^{3} t - 8 s t^{3}}{s^{4} + 2 s^{2} t^{2} + t^{4}}\right) $ (up to sign and swapping $x,y$)
Can the rest of the points be parameterized (or at least some of them)? (such points exist).
What software can do this?
The surface is related to Euler bricks.
EDIT
Some empirical observation about the points of interest. Of the 280 points found 246 have denominator of x that is sum of two squares, 250 have denominator of y that is sum of two squares and 252 have denominator of z that is a square.
Explaining the relation with Euler bricks per John's request (almost sure this have been done before).
So I naïvely wasted my time with perfect Euler bricks. $\begin{align*}a^2+b^2&=s_1^2\qquad \qquad \rm(1)\\ a^2+c^2&=s_2^2\qquad \qquad \rm(2)\\ b^2+c^2&=s_3^2\qquad \qquad \rm(3)\\ a^2+b^2+c^2&=s_4^2\qquad \qquad \rm(4)\end{align*}$
(4) can be parameterized by $(a,b,c) = (2s,2t,1-s^2-t^2)$ To eliminate (1) further parameterize (1) with $(s,t) = (x^2-y^2,2xy)$. Substituting leads to $a=2 x^{2} - 2 y^{2}, b=4 x y ,c= -x^{4} - 2 x^{2} y^{2} - y^{4} + 1$ and leaves only equations (2) and (3). (3) is the surface in question with $z=s_3$. The known parameterization make one of $a,b,c$ $0$.
Without the restricting factor of $2$ in $2s,2t$ the surface becomes
$ (x^{2} + y^{2} - 2 x - 2 y + 2) \cdot (x^{2} + y^{2} - 2 x + 2 y + 2) \cdot (x^{2} + y^{2} + 2 x - 2 y + 2) \cdot (x^{2} + y^{2} + 2 x + 2 y + 2) - z^2=0$
with known solutions $(t,t)$ and $x^2+y^2=2$ (up to sign)