Solution:

$n < 1000$, so the sum of the digits is at most $27$, so $n^{2}$ is a cube not exceeding $27^{3}$. So we are looking for $m^{3}$ which is also a square. That implies $m$ is a square. So the only possibilities are $m = 1, 4, 9, 16, 25$. Giving $n = 1, 8, 27, 64, 125$. The corresponding cubes of the digit sums are $1, 512, 729, 1000, 512$, whereas the corresponding squares are $1, 64, 729, 4096, 15625$. Thus the only solutions are $n = 1, 27$.