Transformations between Pell[-like] equations I’m looking for [non-trivial] transformations that take a Pell-like equation
$$
u^2-dv^2=w
$$
and turn it into another Pell-like equation
$$
x^2-my^2 = z.
$$
Best-case scenario, one could always use the same transformation for a given [target] $m$; but any interesting transformations would be helpful.
EXAMPLE: Let's say I have integers $u,v,w$ such that
$$u^2-6v^2=w.$$
I would like to find functions $f(u,v,w)$, $g(u,v,w)$, and $h(u,v,w)$ such that $x=f(u,v,w)$, $y=g(u,v,w)$, and $z=h(u,v,w)$, and
$$
x^2 - 2y^2 = z.
$$
I know about the $d \to d$ transformations (e.g., turning the Pell-like $u^2-dv^2= \pm 2$ into the Pell $x^2-dy^2=1$). Now I'm looking for something more substantial — a real method of intra-Pellian transformation.
 A: Knowing that if you can find one solution to a Pell equation you can generate others.
If you could target $m$ then:
$x^2 -my^2 = z$
can be written as
$\displaystyle \frac{x^2}{y^2} - \frac{z}{y^2} = m$
$\displaystyle \left( \frac{x}{y} + \frac{\sqrt{z}}{y} \right) \left( \frac{x}{y} - \frac{\sqrt{z}}{y} \right) = m$
This could provide a method for factoring $m$ by looking for $y|x$ and $z$ is square etc.
If you could solve $u^2-dv^2=w$ then transform it into $x^2-my^2 = z$ this could provide a means to factor one number by factoring another.
If a transform that could target $m$ existed it would have significant consequences.

Copy of my comment to the answer:
Trivially:
$$u^2-dv^2=w \rightarrow u^2-(d-c)v^2=w + cv^2$$
but it changes $d$ and if you can reduce the rhs to $1$ by a $d \to d$ transform then you can solve:
$$u^2-(d-c)v^2=1$$
Admittedly there are very limited $d \to d$ transforms.

I haven't had time to fully test or sanity check this:
For the identical rhs case:
$$ x^2 - my^2 = w$$
$$ u^2 -dv^2 = w$$
Try $v = y$
$$ u^2 -dy^2 = w$$
$$ x^2 -u^2 - (m-d)y^2 = 0$$
$$ x^2 - (m-d)y^2 = u^2$$
From Does the special Pell equation $X^2-dY^2=Z^2$ have a simple general parameterization?
we know that $x^2-dy^2=z^2$ is solved by $x=m^2+dn^2, \quad y=2mn, \quad z=m^2-dn^2$. 
i.e. $z$ solves the Pell like equation with the same $d$ value as $z^2$ does.
So we can write:
$$X^2 - (m-d)Y^2 = u$$
but $u$ is  a variable not a constant.
So select an $X$ and a $Y$ and calculate $u$. 
Then solve for $x$ and $y$ in the square rhs equation $ x^2 - (m-d)y^2 = u^2$.

More sanity checks done:
$$u = X^2 - (m-d)Y^2 \, , \, v = 2XY$$
$$x = X^2 + (m-d)Y^2 \, , \, y = 2XY$$
There should be a $w$ in the equations?
The equations give a relationship between $(u,v)$ and $(x,y)$ but do not solve the equations for $w$.

A: Transform from Pell like equation to a ternary quadratic form:
$$ u^2 -dv^2 = k$$
From factorization
$$ d = c^2 - e^2$$
$$ u^2 -(c^2 - e^2)v^2 = k$$
$$ u^2 +  e^2v^2 -c^2v^2 = k$$
write as
$$ x^2 + y^2 - z^2 = k$$
where $x = u$ , $y = ev$ and $z =cv$.
From Diophantine equation $x^2+y^2-z^2=k$
$$x^2+y^2-z^2=k$$
$$y=y$$
$$x=\frac{y^2-k}{2t}-\frac{t}{2}$$
$$z=\frac{y^2-k}{2t}+\frac{t}{2}$$
