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 $-1$.
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