Using the identity ${}^{36}{C}_{r+1}=\frac{36}{r+1}\cdot {}^{35}{C}_r$, the given equation becomes:

$\frac{36}{r+1}=\frac{6}{k^2-3}\Rightarrow k^2-3=\frac{r+1}{6}\Rightarrow k^2=\frac{r+19}{6}$

With $0\le r\le 35$ (so that ${}^{35}{C}_r$ is defined), $19\le r+19\le 54$, hence $k^2\in [19/6,9]=[3.16,9]$.

The only integer perfect squares in this range are $4$ and $9$, giving $k=\pm 2$ (then $r=5$) and $k=\pm 3$ (then $r=35$).

The valid pairs are $(5,2),(5,-2),(35,3),(35,-3)$ — total $4$ elements.