Prove that the expression has infinitely many solutions 
Does $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$ have infinitely many solutions in positive integers $(m,r)$?
Does  $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$ have infinitely many solutions in positive integers $(m,r)$ where $m \equiv 3 \pmod{4}$?

I wasn't sure how to prove that there are infinitely many solutions, but for both of them we need $48r^2$ to be one less than a perfect square. Thus we need $48r^2 = x^2-1$ and so $x^2-48r^2 = 1$, which has infinitely many solutions, but we also need $m$ to be an integer. How do we ensure that in both cases?

The first has no solutions with positive integer $m$. However, the second case,
$$\begin{aligned}
m&=\dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)\\
&=\dfrac{1}{18}\left(x-1\right)\\
&\equiv {3\pmod{4}}
\end{aligned}$$
has infinitely many given by,
$$x= \frac{ (7+\sqrt{48})^{6k+3}+(7-\sqrt{48})^{6k+3}}{2}=1351,\; 9863382151,\; 72010600134783751\dots$$
$$r = \frac{ (7+\sqrt{48})^{6k+3}-(7-\sqrt{48})^{6k+3}}{2\sqrt{48}}=195,\; 1423656585,\; 10393834843080975\dots$$
