I had another look at this last night. Despite obviously having no two-parameter general solution in the usual sense, on account of the gaps in the ranges of valid real z,
it does have a general rational solution involving four unconstrained parameters. The following is a sketch:

Firstly, three cases arise: (1) $z = -1, 0, 1$ (2) $-1 < z < 0$ (3) $1 < z$

Case 1 forces $x = y = 0$, so nothing further need be said about that.

We can flip between Cases 2, 3 by dividing throughout by $z^4$ and replacing $x, y, z$ by $\frac{x}{z^2}, \frac{y}{z^2}, \frac{-1}{z}$ resp. So it suffices to consider one of these, say Case 3.

As alluded to in my initial question, letting $z = \frac{p}{q}$ for coprime integers $p, q$ and homogenizing, we conclude that $|z|$ and $|z^2 - 1|$ are each expressible as a sum of two squares. So in Case 3, $x = d^2 + e^2$ and $(d^2 + e^2)^2 = 1 + f^2 + g^2$. (Note that the form of this ensures that $1 < d^2 + e^2$ for real $f, g$.)

In view of the well-known general rational solution of $x^2 + y^2 + z^2 = t^2$, namely $x : y : z : t = 2 u : 2 v : u^2 + v^2 - 1 : u^2 + v^2 + 1$, we can then conclude that $d^2 + e^2 = \frac{u^2 + v^2 + 1}{2 u}$

Let $u, v = \frac{U}{W}, \frac{V}{W}$ for integers $U, V, W$ with $gcd(U, V, W) = 1$, and suppose a prime $q \equiv 4 Z + 3$ divides $2 U W$, say $q$ divides $U$. If $q$ also divides $U^2 + V^2 + W^2$, and hence $V^2 + W^2$, then $q$ divides $U, V, W$ contrary to $gcd(U, V, W) = 1$. The same applies if $q$ divides $W$. Thus every such prime dividing $2 U W$ does not divide the numerator and therefore must divide the denominator $2 U W$ to an even power, and therefore must divide $2 \frac{U}{W}$ to an even power. So in summary we must have $u = a^2 + b^2$, and we can write:

$ 2 ((a d + b e)^2 + (a e - b d)^2) = (a^2 + b^2)^2 + v^2 + 1$

So denoting $y, x, c = v, a d + b e + a e - b d, a d + b e - a e + b d$, we obtain finally $x^2 - y^2 = (a^2 + b^2)^2 + 1 - c^2$, so that for some rational $h$ we must have :

$x, y = \frac{h^2 + (a^2 + b^2)^2 + 1 - c^2}{2 h}, \frac{h^2 - (a^2 + b^2)^2 - 1 + c^2}{2 h}$

From the preceding two linear equations involving $x, c$ (which have determinant $- 2 (a^2 + b^2)$, which is non-zero) we can then express $d, e$ as rational functions of $a, b, c, h$ etc.

edit (2013-08-11) :

Couple of points, the first expanding on one aspect of the above, and then some more info on the original equation/surface.

Firstly, from the rational solution of $x^2 + y^2 + z^2 = t^2$ near the start, we could equally well conclude that $d^2 + e^2 = \frac{u^2 + v^2 + 1}{u^2 + v^2 - 1}$, and it might be thought this would lead to a different solution or none. But in view of the following identities, this is in fact equivalent to the equation I gave:

$\begin{array}
2 2 u & = & 2 q r\\
2 v & = & (p^2 + q^2 - 1) r\\
u^2 + v^2 - 1 & = & 2 p r\\
u^2 + v^2 + 1 & = & (p^2 + q^2 + 1) r
\end{array}$

where:

$ r, u, v - 1 = \frac{2}{(p - 1)^2 + q^2}, \frac{2 q}{(p - 1)^2 + q^2}, \frac{2 (p - 1)}{(p - 1)^2 + q^2} $ resp

In other words, a suitable multiple of one parametrization leads to the same parametrization with different values and with the "sum of squares less one" term swapped with one of the "even" terms.

Secondly, the original equation is a special case of a so-called Châtelet surface [ http://en.wikipedia.org/wiki/Ch%C3%A2telet_surface ], and I noticed in interesting looking paper on these appeared on the ArXiv only this week http://arxiv.org/abs/1308.0909 ("Rationality problem of generalized Châtelet surfaces", by Aiichi Yamasaki).