Here is the standard geometric argument: after extracting common factors, you are asking for rational points on the quadric $Q\colon x^2-w^2=z^2-y^2$ in $\mathbb{P}^3$, which is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ (as Matt Young's comment explains). Projection from a point on $Q$ gives a birational map $p:Q\to \mathbb{P}^2$ hence a rational parameterisation of $Q$. Explicitly, projecting from $P_0=(0:1:0:1)$ to the $(w=0)$-plane gives the parameterisation
$$
p : (a:b:c)\in\mathbb{P}^2 \mapsto (2ab:c^2-a^2+b^2:2bc:c^2-a^2-b^2) \in Q
$$
of all the points except for the two lines on $Q$ passing through $P_0$ - i.e. the intersection of $Q$ with the tangent plane at $P_0$, which are $z=w$, $x=\pm y$. Notice also that the points $\{b=0,a\ne\pm c\}$ all go to $P_0$, for the same reason.

The advantage of this method is that there is nothing particularly special about the quadratic form $x^2+y^2-z^2-w^2$ here; the same argument parameterises the rational zeros of any nonsingular quadratic form in 3 or more variables, as soon as it has one non-trivial zero.