Junior Balkan Mathematical Olympiad

Solution:
We will prove that the answer is no. Note that
$$m^{3}=9 n^{4}+170 n^{2}+289=(9 n^{2}+17)(n^{2}+17)$$
If $n$ is odd then $m$ is even, therefore $8 \mid m^{3}$. However,
$$9 n^{4}+170 n^{2}+289 \equiv 9+170+289 \equiv 4 \pmod{8}$$
which leads to a contradiction. If $n$ is a multiple of $17$ then so is $m$ and hence $289$ is a multiple of $17^{3}$, which is absurd. For $n$ even and not multiple of $17$, since
$$\gcd\left(9 n^{2}+17, n^{2}+17\right) \mid 9\left(n^{2}+17\right)-\left(9 n^{2}+17\right)=2^{3} \cdot 17$$
this gcd must be $1$. Therefore $n^{2}+17=a^{3}$ for an odd $a$, so
$$n^{2}+25=(a+2)\left(a^{2}-2 a+4\right)$$
For $a \equiv 1 \pmod{4}$ we have $a+2 \equiv 3 \pmod{4}$, while for $a \equiv 3 \pmod{4}$ we have $a^{2}-2 a+4 \equiv 3 \pmod{4}$. Thus $(a+2)\left(a^{2}-2 a+4\right)$ has a prime divisor of type $4 \ell+3$. As it divides $n^{2}+25$, it has to divide $n$ and $5$, which is absurd.