Is there a Fermat-era proof of Theorem 69 from Dickson's Intro to NT?

In Dickson’s Introduction to the Theory of Numbers (Ch. VI, pp. 91-93), he gives the following [wonderful and wonderfully general] theorem.

Theorem 69: All integral solutions of $$x^2-my^2=zw$$ are given by \begin{align} x &= \pm \rho(e\xi u+f\eta u-f\xi v-g\eta v), & y &= \rho(\xi v+\eta u), \\ z &= \rho(e\xi^2+2f\xi\eta+g\eta^2), & w &= \rho(eu^2-2fuv+gv^2), \end{align} where $\rho,\xi,\eta,u,v$ are arbitrary integers, while $e,f,g$ take only the finite set of integral values such that the forms $$F(\xi,\eta)=e\xi^2+2f\xi\eta+g\eta^2$$ are representative forms, just one being chosen from each class of forms of determinant $$f^2-eg=m.$$

Dickson gives several examples (including $m=-1$), and his proof is remarkably elementary (n.b. Dickson developed several proofs over the years, in increasingly elementary terms.)…

But I would offer that the proof is not quite “Fermat-era”: I highly doubt Fermat thought about “representative forms” or “class[es] of forms of [a] determinant”.

So here's my

QUESTION: Does Theorem 69 have a proof using only Fermat-era math, either in the general case, or in the specific cases that Fermat is known to have played around with regularly (i.e., $m=-1$, $m=-2$, and $m=-3$)? For instance, how could Fermat have proven that, given $$a^2+b^2=(c^2+d^2)e, \qquad a,b,c,d,e \in \mathbb{Z}$$ you can always find integers $e_1$ and $e_2$ such that $e=e_1^2+e_2^2$ and [up to sign] $$(a,b)=(ce_1 \pm de_2, ce_2 \mp de_1) \tag{\star}$$ holds?

I know Fermat claimed to have a proof (by infinite descent) of the existence of the form $e=e_1^2+e_2^2$, and have seen proofs of that fact using only Fermat-era math.

But what about the implication/form given by ($\star$)? Is there a proof that you can always find integers $e_1$ and $e_2$ that satisfy, using only Fermat-era math?

In his Diophantine Analysis [pp. 31–33], Carmichael gives a series of lemmas and a final corollary which essentially duplicates Dickson’s theorem in the case $m=-1$. He even goes so far as to say that the result is “due to Fermat”. So it seems that case, at least, already has a Fermat-era proof.