This post had an answer that involved the equation,
$$x^4+y^4+z^4 =t^2$$
Holding $y,z$ constant, this is then a quartic polynomial in $x$ to be made a square. Using a birational transformation, it can be transformed into an elliptic curve. So given an initial rational point $x_0$, one can find an infinite more.
For example, given the Pythagorean triple $a^2+b^2=c^2$, Fauquembergue's elegant solution is,
$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2\tag1$$
From this, I got a second but not so elegant solution,
$$x = \frac{-2a^4b^4+d^2}{2abd}, \quad y =ac,\quad z =bc\tag2$$
where $d = a^4+a^2b^2+b^4$. Scaling variables gives integer $x,y,z$.
However, I used a short-cut, the tangent method, so I'm not sure if $(2)$ is the next simplest. (The first yields $t$ that is $4$th deg in $a,b$, while the second has $t$ that is $16$-deg.)
Q: Is there a simpler solution $x$ such that $t$ as a polynomial in $a,b$ has deg $4<k<16$?
