(*Edited with more details.*)

I know this is a *really* old question, but this has a nice connection to a problem considered by Euler, what is now known as *Euler bricks*, and I couldn't resist. The OP's equation is equivalent to finding three rationals $a,b,c$ such that,

$$\begin{aligned}
a^2+b^2\; &= u_1^2\\
a^2-c^2\; &= u_2^2\\
b^2-c^2\; &= u_3^2
\end{aligned}\tag1$$

A solution (essentially by Euler) is,

$$a = \frac{s^2+1}{2s},\quad b = \frac{t^2+1}{2t},\quad c = 1\tag2$$

where $s,t$ must satisfy,

$$s^2(t^2+1)^2+t^2(s^2+1)^2 = w^2\tag3$$

and is the equation considered by the OP. For *any* solution to $(1)$ with $c\neq0$, then $s,t$ can be recovered as,

$$s = \frac{a\pm\sqrt{a^2-c^2}}{c} = \frac{a\pm u_2}{c}$$

$$t = \frac{b\pm\sqrt{b^2-c^2}}{c} = \frac{b\pm u_3}{c}$$

A small solution to $(3)$ is,

$$s = \frac{4p}{p^2-1},\quad t = \frac{3p^2+1}{p(p^2+3)}\tag4$$

This is the essentially the same one given by Euler for *Euler bricks*,

$$\begin{aligned}
\alpha^2+\beta^2\; &= v_1^2\\
\alpha^2+\gamma^2\; &= v_2^2\\
\beta^2+\gamma^2\; &= v_3^2
\end{aligned}$$

where we have used $p \to p\sqrt{-1}$, and have correspondingly tweaked $(1)-(4)$. From this initial rational point $(4)$, one can then generate an infinite more.

Lectures on Elliptic Curves, or computer algebra systems will do it, to get the rational isomorphism. I suspect we can show that the Mordell-Weil group over $Q(t)$ is only $(Z/2)^2\oplus Z$ by descent or cohomology, but the details maybe tricky, especially if by hand. Another approach, is noting the rank 1 specializations and use Silverman's theorem, III.11 in hisAdvanced Topics in the Arithmetic of Elliptic Curves, effectively. Finally, Rankincreaseupon specialization is difficult, generally. – Junkie May 18 '11 at 2:07