The answer is $181$.
Let $n^2 = (m+1)^3 - m^3 = 3m^2 + 3m + 1$. This implies
$$(2n - 1)(2n + 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.$$
As $(2n - 1, 2n + 1) = (2n - 1, 2) = 1$, one of $2n - 1$ and $2n + 1$ is a square and the other is $3$ times a square.

* If $2n + 1$ is a square, then $3 \mid 2n - 1$, and hence $n \equiv 2 \pmod 3$. But then $2n + 1 \equiv 2 \pmod 3$, so $2n + 1$ cannot be a square. This is a contradiction.

* If $2n - 1$ is a square, let $2n - 1 = a^2$ and $2n + 79 = b^2$. This implies
$$80 = b^2 - a^2 = (b - a)(b + a).$$
To maximize $n$, we need to maximize $b + a$. Since $b - a$ and $b + a$ have the same parity, the maximal case is $(b - a, b + a) = (2, 40)$. In that case, we have $(a, b, n) = (19, 21, 181)$. This means the largest $n$ is $181$.