In Carmichael's Diophantine Analysis ($\S8$), he notes that the equation $$X^2-dY^2=Z^2 \qquad(\dagger)$$ has a two-parameter solution $$x=m^2+dn^2, \quad y=2mn, \quad z=m^2-dn^2. \qquad(\star)$$ He then says “there is no ready means for determining whether this is the general solution”, and goes on to give a [much more complicated] general solution.
Is Carmichael’s conclusion true? i.e. Can it neither be proven nor disproven [easily] that $(\star)$ is indeed a general parametric solution to $(\dagger)$?
EDIT: In Barbeau’s Pell’s Equation (pg. 36), it says:
Exercise 2.8. [...] (a) Consider the case d=2 [of the equation $U^2-W^2=dV^2$]. Obtain the parametric solutions $$(u,v,w)=(r^2+2s^2,2rs,2s^2-r^2)$$ and $$(u,v,w)=(2r^2+s^2,2rs,s^2-2r^2).$$ (b) Obtain a parametric set of solutions for $u^2-dv^2=w^2$.
Given that this ‘exercise’ appears early in such an elementary textbook, it seems likely to me that there is a solution similar to $(\star)$ which is valid [though clearly $(\star)$ is not, as pointed out in the comments].
EDIT: Maybe Barbeau means let $d=d_1d_2$ and $v=v_1v_2$ be arbitrary factorizations, so that $dv^2=u^2-w^2=(u-w)(u+w)$ implies $u-w=d_1v_1^2$ and $u+w=d_2v_2^2$, yielding the parametric solution $$(u,v,w) = \biggl(\frac{d_2v_2^2+d_1v_1^2}{2},v_1v_2,\frac{d_2v_2^2-d_1v_1^2}{2}\biggr).$$
Actually, the complete rational solution to,
$$x^2+ny^2 = z^2$$
needs a third parameter, the scaling factor $t$, hence,
$$\big((a^2-nb^2)t\big)^2+n(abt)^2 = \big((a^2+nb^2)t\big)^2$$
Proof of completeness:
Let $xyz \ne 0$. Using a system of three equations,
$$\begin{aligned} x\,&=(a^2-nb^2)t\\ y\,&=abt\\ z\,&=(a^2+nb^2)t \end{aligned}$$
one can then rationally express the three unknowns $a,b,t$ in terms of the knowns $x,y,z$ by the simple formulas,
$$\begin{aligned} a\,&=x+z\\ b\,&=y\\ c\,&=\tfrac{1}{2(x+z)} \end{aligned}$$
though one needs a little algebraic manipulation to get these forms.
P.S. Just a minor quibble, $x^2-dy^2 =N$ is a Pell-like equation, while $ax^2+by^2+cz^2 = 0$ is a ternary quadratic form.