jmc

Write $n^2=(m+1{)}^3-m^3=3m^2+3m+1$, or equivalently, $(2n+1)(2n-1)=4n^2-1=12m^2+12m+3=3(2m+1{)}^2$. Since $2n+1$ and $2n-1$ are both odd and their difference is $2$, they are relatively prime. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have $2n-1$ be three times a square, for then $2n+1$ would be a square congruent to $2$ modulo $3$, which is impossible. Thus $2n-1$ is a square, say $b^2$. But $2n+79$ is also a square, say $a^2$. Then $(a+b)(a-b)=a^2-b^2=80$. Since $a+b$ and $a-b$ have the same parity and their product is even, they are both even. To maximize $n$, it suffices to maximize $2b=(a+b)-(a-b)$ and check that this yields an integral value for $m$. This occurs when $a+b=40$ and $a-b=2$, that is, when $a=21$ and $b=19$. This yields $n=181$ and $m=104$, so the answer is $\boxed{181}$.