Suppose that $ m^3$ is the sum of $ m$ squares $ (n+1)^2,\ (n+2)^2,\ \dots,\ (n+m)^2$. Then \[ m^3 = \sum_{i=1}^m (n+i)^2 = \sum_{i=1}^{m+n} i^2 - \sum_{i=1}^n i^2 = \frac{(m+n)(m+n+1)(2m+2n+1)}{6} - \frac{n(n+1)(2n+1)}{6} = \frac{2m^3 + (6n + 3) m^2 + (6n^2 + 6n + 1) m}{6}.\] Hence, we need to solve \[ n^2 + (m+1)n - \frac{4m^2 - 3m - 1}{6} = 0\] in positive integers. The discriminant of this quadratic equation w.r.t. $ n$ is $ \frac{11 m^2 + 1}{3}$ and it must be a square of some integer $ q$. Positive solutions to this generalized Pell equation $ 11 m^2 + 1=3q^2$ is given by the following recurrent sequence: $ m_0 = 1,\ m_1 = 47,\ m_{k+1} = 46 m_k - m_{k-1}$ which can be alternatively described as \[ m_k = \frac{11+2\sqrt{33}}{22}(23+4\sqrt{33})^k + \frac{11-2\sqrt{33}}{22}(23-4\sqrt{33})^k.\] Corresponding values of $ n$ are given by $ n_0 = 0,\ n_1 = 21,\ n_{k+1} = 46 n_k - n_{k-1} + 22$.