I. Near-perfect Euler bricks
Given System 1,
\begin{align} a^2 + b^2 &= \square_1\\[4pt] a^2 + c^2 &= \square_2\\[4pt] b^2 + c^2 &= \square_3 + \color{red}{h^2}\\[4pt] a^2 + b^2 + c^2 &= \square_4 + \color{red}{h^2}\end{align}
If $h=0$, then this is the perfect Euler brick, though none have been found. System 1 is equivalent to System 2,
\begin{align} a^2 + b^2 &= \square_a\\[4pt] a^2 + c^2 &= \square_b\\[4pt] a^2 + d^2 &= \square_c\\[4pt] b^2 + c^2 - d^2 &= h^2\end{align}
In this post, Tomita (after some modifications I made) found a solution $(a,b,c,d)$ to the first three equations as,
\begin{align} a &= 2(n^2-1)\\[3pt] b &= 3(n^2-1)/2\\[3pt] c &= \frac{(n^2-1)(z^2-1)}z\\[3pt] d &= \frac{(n^2-1)^2-z^2}z\end{align}
where the fourth becomes a quartic in $z$ to be made a square,
$$\frac{4 n^2 (n^2 - 2) z^4 + 9 (n^2 - 1)^2 z^2 - 4 n^2 (n^2 - 2) (n^2 - 1)^2}{4z^2} = h^2$$
I found this has the simple initial polynomial solution,
$$z_1 = \frac{n^2-1}n\qquad$$
II. Question
Q: Since the quartic in $z$ is birationally equivalent to an elliptic curve, then there are infinitely many rational $z_n$. What is the simplest polynomial $z_2, z_3,$ and so on?
