Solutions to $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$ 
Find all positive integer solution pairs $(m,r)$ to $m = \dfrac{1}{18}\left(\sqrt{48r^2+1}-1\right)$.

I thought about using Pell's equation, but that only tells us solutions such that $\sqrt{48r^2+1}$ is an integer. How do we deal with the $\dfrac{1}{18}$ and $-1$?
 A: We can proceed similarly as in my answer to your previous question, though this time it's to find the solutions. $(18m+1)^2=48r^2+1$ simplifies to $$3m(9m+1)=4r^2,$$so $r^2=9k^2$, $m=3n$ for some $k,n$. It follows that $$n(27n+1)=4k^2,$$ and since the factors of the LHS are coprime, both of them must be squares, say $n=y^2, 27n+1=x^2$. Therefore, $$x^2-27y^2=x^2-3(3y)^2=1.$$To find the fundamental solution $(x_0,y_0)$ we need to go through the convergents of the continued fraction of $\sqrt{3},$ but we're lucky because the $3y$ implies that the denominator of the "solving convergent" is a multiple of $3$: we quickly find the fifth convergent $\frac{26}{15}$ to provide us with $(x_0,y_0)=(26,5)$. As a consequence, $m=3\cdot5^2=75$ and $r=13\cdot15=195$ are solutions of your equation; defining $\{z_n\}$ to be the sequence of the indexes of the convergents that have a denominator divisible by $3$, all the other solutions $(m_n,r_n)$ can be deduced by the solutions of $x^2-3(3y)^2=1$, i.e.$$x_{n}=\frac{(2-\sqrt3)^{z_n}+(2+\sqrt3)^{z_n}}{2}$$and $$y_n=\frac{(2+\sqrt3)^{z_n}-(2-\sqrt3)^{z_n}}{6\sqrt3}.$$
