For any integer $m > 1$ the numbers $2m^2 + 1$, $m(2m^2 + 1)$, $-m(2m^2 + 1)$ satisfy the conditions of the problem, because they are pairwise different and
$$\begin{aligned} & (2m^2 + 1)^2 + (m(2m^2 + 1))^2 + (-m(2m^2 + 1))^2 \\ &= (1 + m^2 + m^2) \cdot (2m^2 + 1)^2 = (2m^2 + 1)^3 = (1 + m^3 - m^3) \cdot (2m^2 + 1)^3 \\ &= (2m^2 + 1)^3 + (m(2m^2 + 1))^3 + (-m(2m^2 + 1))^3. \end{aligned}$$
With $m$ growing, the numbers in these triples get arbitrarily large, hence for any set of these triples one can find a new triple, where all numbers are larger than the ones already used.
Any positive integer $k$ can be written as $k = 3q + r$ with $0 \le r < 3$. Choose $q$ triples as above so that the numbers in them do not coincide. If $r = 1$, then add $0$, and if $r = 2$, then add $0$ and $1$. Since for each group the sum of the squares of the numbers equals the sum of the cubes of the numbers, the same property holds for the whole set.