Variation of Pell's equation Is it possible to find all integer solutions of the Diophantine equation
$$
2x^{2} - 3y^{2} = 5
$$
I think we have to use $\mathbb{Q}(\sqrt{6})$ somewhere, but I don't know how to use units of $\mathcal{O}_{\mathbb{Q}(\sqrt{6})}$. 

I just found a way to find solutions, and I think these are all: 
If $2x^{2} - 3y^{2} = 5$ and $z^{2} - 6w^{2} = 1$, then we have 
$$
2(xz + 3yw)^{2} - 3(yz + 2xw)^{2} = (2x^{2} - 3y^{2})(z^{2} - 6w^{2}) = 5
$$
which follows from $N(\sqrt{2}x + \sqrt{3}y) = 5$ and $N(z - \sqrt{6}y) = 1$. Here the norm map $N$ is $N:\mathbb{Q}(\sqrt{6})\to \mathbb{Q}$. 
Since we know a solution for the original equation $(x, y) = (2, 1)$ and the second equation $(5, 2)$, and those are all since "quotient" of two solutions will be a unit in $\mathcal{O}_{\mathbb{Q}(\sqrt{6})}$, which is generated by the fundamental unit $5 + 2\sqrt{6}$ and $-1$. 
 A: The form $f(X,Y) = 2X^2-3Y^2$ is a primitive integral indefinite binary quadratic form. The stabilizer of the action of the modular group $\mathrm{PSL}(2,\mathbf{Z})$  is generated by the matrix $W = \begin{pmatrix} 5&6\\4&5\end{pmatrix}$ . Therefore, for any integer $n$, the pair $W^n\begin{pmatrix} 2\\1\end{pmatrix}$ is a solution to your equation.
In general, when one begins with a primitive integral indefinite binary quadratic form and wants to solve the equation $f = m$ for some integer $m$, if there is one solution to this equation, then there are infinitely many such solutions. What one does is to compute a generator of the stabilizer (which is isomorphic to $\mathbb{Z}$) and use this generator along with a known solution to produce infinitely many solutions.
There is a software -- still being developed -- but needs some pre-reading.
Here are relevant links:
software : http://math.gsu.edu.tr/azeytin/infomod/node/3
book : Buchmann & Vollmer, Binary quadratic forms
article : http://math.gsu.edu.tr/azeytin/pdfs/bqf.pdf
A: ADDED: for your question $2x^2 - 3 y^2 = 5,$ we get two orbits, each can be made into a linear recursion by Cayley-Hamilton:
$$ x= 2, 16, 158, 1564, 15482, ... $$
$$ y = 1, 13, 129, 1277, 12641, ...     $$
with
$$ x_{n+2} = 10 x_{n+1} - x_n , $$
$$ y_{n+2} = 10 y_{n+1} - y_n . $$
The second orbit begins
$$ x= 4, 38, 376, 3722, 36844, ... $$
$$ y = 3, 31, 307, 3039, 30083, ...     $$
with
$$ x_{n+2} = 10 x_{n+1} - x_n , $$
$$ y_{n+2} = 10 y_{n+1} - y_n . $$
ORIGINAL:Well, I don't have a diagram for $2x^2 - 3 y^2$ ready, maybe i could do that later. I do have one for $3 x^2 - 2 y^2,$ and you may read what you require as $3x^2 - 2 y^2 = -5$ but then switch the variables.
Note that the generator matrix for the oriented automorphism group is visible in the pair regions highlighted with asterisks * above and below, where the original form values 3,-2 are repeated.  In Weissman's recent book, he would say that I draw both the domain topograph and the range topograph together. The domain is the green column vectors, the range is the pink values. 
 
