This problem was proposed by Bulgaria

We can get to $m^3=(9n^2+17)(n^2+17)$ easily. If $17|n$, we will have $17^3 | m^3$ which will result in a contradiction. Hence, we have $17$ and $n$ are coprime. Let $ \text{gcd}(9n^2+17, n^2+17)=d$. It follows that $d | 8 \cdot 17 \Rightarrow d|8$. If $d=1$, assume we have $9n^2+17=a^3$ and $n^2+17=b^3$ for positive integers $a$ and $b$. $\text{mod 7}$ will give a clear contradiction. If $d=2$, we will have $8|(9n^2+17)(n^2+17)$ which is again a clear contradiction because of $\text{mod 4}$ If $d=4$ or $d=8$, and we will get to the contradiction by $ \text{mod 4}$

A slightly harder but conceptually good way to deal with $n^2 + 17 = b^3$ for $d=1$ (i.e. even $n$, odd $b$, and ignoring the other equation) is to write it as $n^2 + 5^2 = (b+2)(b^2-2b+4)$ and note that the right hand side always has a factor $\equiv 3 \pmod 4$, regardless of whether $b\equiv 1 \pmod 4$ or $b\equiv 3 \pmod 4$; however this is impossible for the left-hand side.