Write $\frac{n^{2}+7 n+136}{n-1}=n+\frac{8 n+136}{n-1}=n+8+\frac{144}{n-1}=9+(n-1)+\frac{144}{(n-1)}$. We seek to find $p$ and $q$ such that $p q=144$ and $p+q+9=k^{2}$. The possibilities are seen to be $1+144+9=154,2+72+9=83,3+48+9=60,4+36+9=49,6+24+9=39$, $8+18+9=35,9+16+9=34$, and $12+12+9=33$. Of these, $\{p, q\}=\{4,36\}$ is the only solution to both equations. Hence $n-1=4,36$ and $n=5,37$.