Infinitely many triangular numbers that are squares

Reduction. $\frac{n(n+1)}2=m^2\iff 4n^2+4n=8m^2\iff(2n+1)^2-8m^2=1$. So it suffices to find infinitely many solutions $(x,y)$ of $x^2-8y^2=1$ in positive integers with $x$ odd: then $n=\frac{x-1}2$ and $m=y$.

Generating solutions. Suppose $x^2-8y^2=1$. Put $x'=3x+8y$, $y'=x+3y$. Then
$$x'^2-8y'^2=9x^2+48xy+64y^2-8x^2-48xy-72y^2=x^2-8y^2=1.$$
Also $x'\equiv x\pmod 2$ (since $8y$ is even), so $x'$ is odd if $x$ is, and $x'>x$, $y'>y$.

Start. $(x,y)=(3,1)$ satisfies $9-8=1$ and $x$ is odd. The recurrence gives $(3,1)\to(17,6)\to(99,35)\to(577,204)\to\cdots$, an infinite strictly increasing sequence of solutions with $x$ odd.

Result. The corresponding $n=\frac{x-1}2$ are $1,8,49,288,\dots$, with $T_1=1^2$, $T_8=36=6^2$, $T_{49}=1225=35^2$, $T_{288}=41616=204^2$.

Remark. Pell's theory shows these are *all* the solutions (they come from powers of $3+2\sqrt2$).

Lesson. A Pell-type equation has an infinite family generated by multiplying by a unit such as $3+\sqrt8$; reduce a problem to $x^2-Dy^2=1$ by completing the square.