The given equation is equivalent to $(x-2y)^2 - 3y^2 = 1$.

Substituting $u=x-2y$ leads to the following Pell's equation:

$$ u^2 - 3y^2 = 1$$

Its fundamental solution is $(u_1, y_1) = (2, 1)$. Hence all its solutions are given by:

$$ \left\{ \begin{array}{ll} u_n &= \frac{(2 + \sqrt{3})^n + (2-\sqrt{3})^n}{2}\\ \\ y_n &= \frac{(2 + \sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}\\ \end{array} \right. $$

Accordingly the solutions to the original equations are given by:

$$ \left\{ \begin{array}{ll} x_n &= \frac{(2 + \sqrt{3})^n + (2-\sqrt{3})^n}{2} +\frac{(2 + \sqrt{3})^n - (2-\sqrt{3})^n}{\sqrt{3}}\\ \\ y_n &= \frac{(2 + \sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}\\ \end{array} \right. $$