An annoying Pell-like equation related to a binary quadratic form problem Let $A,B,C,D$ be integers such that $AD-BC= 1 $ and $ A+D = -1 $.

Show by elementary means that the Diophantine equation
  $$\bigl[2Bx + (D-A) y\bigr] ^ 2 + 3y^2 = 4|B|$$
has an integer solution (that is, a solution $(x,y)\in\mathbb Z^2$). If possible, find an explicit solution (involving $A,B,C,D$, of course).

Motivation: I arrived at this equation after trying to find explicitly the matrix $g$ suggested by Will Jagy on his answer to this question of mine. Concretely, if $\gamma=\binom{A\ \ B}{C\ \ D}$, then $\gamma$ has order $3$ in $\operatorname{SL_2}(\mathbb Z)$. By indirect methods it can be shown that $\gamma$ is conjugated in $\operatorname{SL_2}(\mathbb Z)$ to one of the matrices $P$ or $P^{-1}$, being $P=\binom{\ \ \,0\quad1}{-1\ \ -1}$ (see studiosus' answer to the same question.). Unfortunately this argument is rather sophisticated to my knowledge, and besides I think that a direct argument is possible.
Because of this I tried to find a explicit matrix $g=\binom{x\ \ y}{z\ \ w}\in\operatorname{SL_2}(\mathbb Z)$ such that $gP=\gamma g$ or $gP^{-1}=\gamma g$. The matricial equalities lead to a system of $4$ linear equations in the unknowns $x,y,z,w$ , which can be easily solved. Plugging these solutions $(x,y,z,w)$ (recall that we are considering the two possibilities of conjugation, to $P$ or $P^{-1}$) into the equation $xy-zw=1$ yields $Bx^2+(D-A)xy+(-C)y^2=\pm1$. Completing the square and using the equalities $AD-BC=1$ and $A+D=-1$ we obtain the required equation. I tried to solve it explicitly, with no success.

You have the binary quadratic form
$$ \color{red}{ f(x,y) = B x^2 + (D-A)xy - C y^2}  $$ in your last paragraph. The discriminant is $$ \Delta = (D-A)^2 + 4 B C.  $$
You also have $AD-BC = 1$ and $A+D = -1.$ So, $BC - AD = -1$ and
$$ A^2 + 2 AD + D^2 = 1,  $$
$$  4BC - 4AD = -4, $$
$$ A^2 - 2 AD + D^2 +4BC = -3,  $$
$$  \Delta =  (A -  D)^2 +4BC = -3.  $$
If we had $\gcd(A-D,B,C) > 1$ we would have a square factor of $\Delta,$ so that is out, the coefficients of $f(x,y)$ are relatively prime (as a triple, not necessarily in pairs). Next, $-3$ is not a square, so we cannot have $B=0$ or $C=0.$ 
Finally $f$ is definite. If, say, $B > 0,$ it is positive definite. With discriminant $-3,$ it is then equivalent in $SL_2 \mathbb Z$ to
$$g(u,v) = u^2 + u v + v^2$$ and both $g$ and $f$ integrally represent $1.$ 
If $B < 0,$ then $f$ is negative definite. With discriminant $-3,$ it is then equivalent in $SL_2 \mathbb Z$ to
$$h(u,v) = -u^2 - u v - v^2$$ and both $h$ and $f$ integrally represent $-1.$ 
That is enough for what you asked. I should point out that the reduction for positive binary forms is very similar to the Euclidean algorithm for finding GCD is very similar to the algorithm for finding the matrix $W$ in $SL_2 \mathbb Z$ that is conjugate to your original and has, for example, the smallest value of 
$(w_{22} - w_{11})^2.$ I recommend Buell, Binary Quadratic Forms if I have not already. I was going to answer with a reduction that minimized $(A-D)^2,$ then looked again at your question and realized there was a shortcut. Note that many books do reduction for positive definite binary forms, and give short lists for discriminants of small absolute value.
