Find all integers $m>1$ such that $m^3$ is a sum of $m$ squares of consecutive integers.
1. **Assume the form of the sum of squares:**
   Suppose that \( m^3 \) is the sum of \( m \) squares of consecutive integers. Let these integers be \( n+1, n+2, \ldots, n+m \). Then we have:
   \[
   m^3 = \sum_{i=1}^m (n+i)^2
   \]

2. **Express the sum of squares:**
   The sum of squares of the first \( k \) integers is given by:
   \[
   \sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}
   \]
   Therefore, the sum of squares from \( n+1 \) to \( n+m \) can be written as:
   \[
   \sum_{i=1}^m (n+i)^2 = \sum_{i=1}^{m+n} i^2 - \sum_{i=1}^n i^2
   \]

3. **Substitute the sum of squares formula:**
   \[
   \sum_{i=1}^{m+n} i^2 = \frac{(m+n)(m+n+1)(2m+2n+1)}{6}
   \]
   \[
   \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   Thus,
   \[
   \sum_{i=1}^m (n+i)^2 = \frac{(m+n)(m+n+1)(2m+2n+1)}{6} - \frac{n(n+1)(2n+1)}{6}
   \]

4. **Simplify the expression:**
   \[
   m^3 = \frac{(m+n)(m+n+1)(2m+2n+1) - n(n+1)(2n+1)}{6}
   \]
   Expanding and simplifying the numerator:
   \[
   (m+n)(m+n+1)(2m+2n+1) = 2m^3 + (6n + 3)m^2 + (6n^2 + 6n + 1)m
   \]
   \[
   n(n+1)(2n+1) = 2n^3 + 3n^2 + n
   \]
   Therefore,
   \[
   m^3 = \frac{2m^3 + (6n + 3)m^2 + (6n^2 + 6n + 1)m - (2n^3 + 3n^2 + n)}{6}
   \]

5. **Set up the equation:**
   For \( m^3 \) to be an integer, the numerator must be divisible by 6. This leads to:
   \[
   6m^3 = 2m^3 + (6n + 3)m^2 + (6n^2 + 6n + 1)m - (2n^3 + 3n^2 + n)
   \]
   Simplifying, we get:
   \[
   4m^3 = (6n + 3)m^2 + (6n^2 + 6n + 1)m - (2n^3 + 3n^2 + n)
   \]

6. **Solve the quadratic equation:**
   We need to solve:
   \[
   n^2 + (m+1)n - \frac{4m^2 - 3m - 1}{6} = 0
   \]
   The discriminant of this quadratic equation with respect to \( n \) is:
   \[
   \Delta = (m+1)^2 + \frac{4(4m^2 - 3m - 1)}{6}
   \]
   Simplifying the discriminant:
   \[
   \Delta = m^2 + 2m + 1 + \frac{16m^2 - 12m - 4}{6}
   \]
   \[
   \Delta = m^2 + 2m + 1 + \frac{8m^2 - 6m - 2}{3}
   \]
   \[
   \Delta = m^2 + 2m + 1 + \frac{8m^2 - 6m - 2}{3}
   \]
   \[
   \Delta = \frac{11m^2 + 1}{3}
   \]
   For \( \Delta \) to be a perfect square, we need:
   \[
   11m^2 + 1 = 3q^2
   \]

7. **Solve the Pell's equation:**
   The positive solutions to the generalized Pell's equation \( 11m^2 + 1 = 3q^2 \) are given by the sequence:
   \[
   m_0 = 1, \quad m_1 = 47, \quad m_{k+1} = 46m_k - m_{k-1}
   \]
   Alternatively,
   \[
   m_k = \frac{11 + 2\sqrt{33}}{22}(23 + 4\sqrt{33})^k + \frac{11 - 2\sqrt{33}}{22}(23 - 4\sqrt{33})^k
   \]

8. **Find corresponding values of \( n \):**
   The corresponding values of \( n \) are given by:
   \[
   n_0 = 0, \quad n_1 = 21, \quad n_{k+1} = 46n_k - n_{k-1} + 22
   \]

The final answer is \( \boxed{ m = 1, 47, \ldots } \)