Find all triangular numbers that are perfect squares (recall that the $n$-th triangular number is $1+2+\ldots+n$).
We are looking for integers that can be written in the form $\frac{n(n+1)}{2}$ for some $n$, and such that there exists $y \in \mathbb{N}^{*}$ satisfying $\frac{n(n+1)}{2}=y^{2}$, which is equivalent to $(2 n+1)^{2}-8 y^{2}=1$. The Pell equation $X^{2}-8 Y^{2}=1$ has $(3,1)$ as its minimal solution, and thus the first coordinate of the general solutions is written as

$$
x_{k}=\frac{1}{2}\left((3+2 \sqrt{2})^{k}+(3-2 \sqrt{2})^{k}\right)=\frac{1}{2}\left((1+\sqrt{2})^{2 k}+(1-\sqrt{2})^{2 k}\right)
$$

for all $k \geqslant 0$. The integers $n$ corresponding to these are those of the form $\frac{x_{k}-1}{2}$ (note that $x_{k}$ is necessarily odd for all $k$), so the numbers we are looking for are the $\frac{x_{k}^{2}-1}{8}$.