About the quartic Diophantine equation:
$$ 2 x^4 - 2 x^2 = 3 (y^2 - 1)$$
On oeis.org/A180445 it says that all positive solutions $(x,y)$ are: $$(1,1)\ \ (2,3)\ \ (3,7) \ \ (6,29)\ \ (91,6761)$$
I am looking for a simple proof for that (if any simple proof exists, because I know that these kind of equations can be very difficult). Introducing $X=2x^2-1$, it becomes
$$6 y^2-X^2 = 5$$ which may be more tractable.
EDIT: Indeed, here the solutions are infinitely many: there are two families, for all $k\ge 1$. Let,
$$u_1 =(5 - 2 \sqrt6)^k\\ u_2 = (5 + 2 \sqrt6)^k$$
then,
$$X_k=\frac{ ( 1 + \sqrt6) u_1 +(1 - \sqrt6) u_2}{2}$$ $$y_k=\frac{ (6 + \sqrt6) u_1 + (6 - \sqrt6) u_2}{12}$$ and $$X_k=\frac{ (1 - \sqrt6) u_1 + (1 + \sqrt6) u_2}{2}$$ $$y_k=\frac{ (6 - \sqrt6) u_1+ (6+ \sqrt6) u_2}{12}$$
but still I don't know how to sort out those corresponding to solutions to the original equation.