Starting with $Z(Z - m^2) = (Z - n^2) b^2$, to which Allan MacLeod's elliptic curve can be reduced by taking $W = (Z - n^2)b$, one can find a general parametrization of m, n as follows.
Letting $a, x = Z/b, m/b$ gives $a^2 - a b x^2 = a b - n^2$, in which then letting $n = a y$ gives $b (x^2 + 1) = a (y^2 + 1)$. This implies $x^2 + 1, y^2 + 1 = a z, b z$ for some rational $z$, and multiplying these gives after composition $(\frac{x y + 1}{z})^2 + (\frac{x - y}{z})^2 = a b$.
Letting $a, b = k A, k B$, where $A, B$ are coprime integers, the factor $k^2$ in the preceding equation can be absorbed into each square, and we can conclude that $A, B$ are each a sum of two squares, say $p^2 + q^2, r^2 + s^2$ resp, so that $B (x^2 + 1) = A (y^2 + 1)$ becomes $(p x + q)^2 + (p - q x)^2 = (r y + s)^2 + (r - s y)^2$
The latter has general solution as follows, for rational $u, v$ with $u^2 + v^2 = 1$ :
$p x + q, p - q x = u (r y + s) + v (r - s y), v (r y + s) - u (r - s y) w$
So that:
$ x = \frac{u (r y + s) + v (r - s y) - q}{p} = \frac{p - v (r y + s) - u (r - s y)}{q}$
which expresses $y$ rationally in terms of $u, v$ and $p, q, r, s$ (and the latter appear homogenously, so one of them is disposable i.e. can be assumed equal to 1).