I am looking for cases in which $\sqrt{3n(n+1)}$ is an integer, i.e. cases in which
$$
3n(n+1)=m^2,\quad m\in\mathbb{N}.
$$
I can find solutions such as
$$
n=0,3,48,675,9408,131043,\dots
$$
and I expect this list to be infinite. Is it? Is there a straightforward way to prove these kinds of statements?

There are many ways of reframing the problem: finding integers $n$ that are simultaneously 3 times a perfect square and 1 less than another perfect square (choosing $3n$ and $n+1$ to each be perfect squares), etc. Then I could set $n=3k^2$ and try to solve for cases in which
$$
3k^2+1=l^2 \quad\Leftrightarrow\quad 3k^2=(l+1)(l-1)\quad k,l\in\mathbb{N}.
$$
It seems plausible that there are infinite solutions given the various formulas for perfect squares but none of the rabbit holes that I followed led me anywhere productive.