This is more of a comment.
If I understand correctly, your Super-4 a perfect cuboid will give a perfect cuboidSuper-4 according to On Perfect Cuboids
Are there four squares all pairs of which have square differences? For a perfect cuboid we could take the squares of $y_3 z$, $y_2 y_3$ , $x_1 z$ and $x_1 y_3$.
While wasting my time with perfect cuboids, I found 2 surfaces on which they are nontrivial rational points. The first might be for all perfect cuboids, the second is not for all. The surfaces are:
$$ x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} - 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 4 x^{2} y^{2} z^{4} + x^{4} y^{2} + x^{2} y^{4} - 8 x^{2} y^{2} z^{2} + x^{2} z^{4} + y^{2} z^{4} - 4 x^{2} y^{2} - 2 x^{2} z^{2} - 2 y^{2} z^{2} + x^{2} + y^{2} = 0$$
and
$$ x^{4} y^{3} z^{3} - x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} + 2 x^{4} y^{3} z + 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 2 x^{4} y z^{3} + 4 x^{2} y^{3} z^{3} - x^{4} y^{2} + x^{2} y^{4} - 2 x^{4} y z + 4 x^{2} y^{3} z - 4 x^{2} y z^{3} + 2 y^{3} z^{3} + x^{2} z^{4} - y^{2} z^{4} - 4 x^{2} y z + 2 y^{3} z - 2 x^{2} z^{2} + 2 y^{2} z^{2} - 2 y z^{3} + x^{2} - y^{2} - 2 y z =0$$
In machine readable form:
x^4*y^2*z^4 + x^2*y^4*z^4 - 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 4*x^2*y^2*z^4 + x^4*y^2 + x^2*y^4 - 8*x^2*y^2*z^2 + x^2*z^4 + y^2*z^4 - 4*x^2*y^2 - 2*x^2*z^2 - 2*y^2*z^2 + x^2 + y^2 = 0
and
2*x^4*y^3*z^3 - x^4*y^2*z^4 + x^2*y^4*z^4 + 2*x^4*y^3*z + 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 2*x^4*y*z^3 + 4*x^2*y^3*z^3 - x^4*y^2 + x^2*y^4 - 2*x^4*y*z + 4*x^2*y^3*z - 4*x^2*y*z^3 + 2*y^3*z^3 + x^2*z^4 - y^2*z^4 - 4*x^2*y*z + 2*y^3*z - 2*x^2*z^2 + 2*y^2*z^2 - 2*y*z^3 + x^2 - y^2 - 2*y*z = 0