Equation if we write in the General form:
$$aX^2+bXY+cY^2=eZ^2+jZW+tW^2$$
If in this equation there any equivalent to a quadratic form in which the root is an integer.
$$q=\sqrt{b^2+4a(e+j+t-c)}$$
Then there are solutions. They can be written by making the replacement.
$$x=(b(2(e+j+t)-b)+4ac)s-(b+2a)(j+2t)k$$
$$y=(b^2+4c(e+j+t-a))s^2-2(b+2c)(j+2t)sk+(j^2+4t(a+b+c-e))k^2$$
Then decisions can be recorded and they are as follows:
$$X=(b-2(e+j+t-c)\pm{q})p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((2(e+j+t-c)-b)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$Y=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(2(e+j+t-a)-b)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(2(e+j+t-a)-b)s)x)n^2$$
$$***$$
$$Z=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$W=(\pm{q}-(b+2a))p^2+2(q((2(a+b+c-e)-j)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((2(a+b+c-e)-j)k-(b+2c)s)x)n^2$$
$p,n,k,s $ - integers asked us.
Given a constant $\, D := 108,\,$ consider the function defined by
$$ f(x,y) := 9 + D\,x^2(1+x) - y^2 \tag{1} $$
where $\, f(x,y) = 0\,$ is the equation of an elliptic curve.
It is equivalent to the curve in Weierstrass form
$$ E: y^2 = x^3 + 1/D\,x^2 + 12/D^3 \tag{2} $$
in the following sense. $\,f(x,y) = 0 \,$ iff $\,[x/D,y/D^2]\,$ is
a point of curve $\,E\,$
which has $j$-invariant equal to $\,-12288/25.\,$
This curve is equivalent to the LMFDB 135.a1 curve
$$ E135a: y^2 + y = x^3 - 3x + 4. \tag{3} $$
The curve $\,E\,$ has a rational point $\,P:=[1/D,15/D^2]\,$ of rank $\,1\,$ which PARI/GP seems not to be able to provide, but it does
provide a rational generating point $\,[4,-8]\,$ for curve $\,E135a\,$.
Given a point $\,[x,y]\,$ of $\,E135a\,$ then
$\,[(x-1)/(3D),(-2y-1)/D^2]\,$ is a point of $\,E.\,$
Each point $\,[x,y]\,$ of $\,E\,$ satisfies
$$ 0 = f(x\, D,y\, D^2). \tag{4} $$
However there are only a finite number of integer solutions to $\,f(x,y)=0.\,$
They correspond to the generator multiples
$$ 1P\mapsto(1,15),\, 2P\mapsto(0,3),\, 4P\mapsto(-1,-3),\, 7P\mapsto(6,-135),\,
8P\mapsto(4,93). \tag{5}$$
Each solution $\,(x,y)\,$ yields a solution $\,(x,-y).\,$ There
are no other solutions in integers.
Some PARI/GP code is
D = 108; E = ellinit([0,1/D,0,0,1/12/D^2]); P = [1/D,15/D^2];
E135a = ellinit([0, 0, 1, -3, 4]);
print("E:",ellgenerators(E));
print("E135a:",ellgenerators(E135a));
for(n=1, 9, print(n," ",[x,y]=ellmul(E,P,n); [x*D,y*D^2]));
Best Answer
It needs to be (5) because (3), (4) imply $(x,\,y)$ is equal to its set of possible values rather than an element of that set.