Croatian Mathematical Olympiad

Let us write the equation in the form
$$5m^3 = n(27n^3 - 2n + 1)$$
and note that $n$ and $27n^3 - 2n + 1$ are relatively prime numbers. We consider two cases:

a) $n = 5a^3$ and $27n^3 - 2n + 1 = b^3$ for some positive integers $a$ and $b$.
Inequality $(3n-1)^3 = 27n^3 - 27n^2 + 9n - 1 < 27n^3 - 2n + 1$ is equivalent to the inequality $27n^2 - 11n + 2 > 0$, which holds for all positive integers $n$. Also, we have $27n^3 - 2n + 1 < 27n^3 - 2 + 1 < 27n^3 = (3n)^3$.
Hence, $(3n-1)^3 < 27n^3 - 2n + 1 < (3n)^3$, which is impossible.

b) $n = a^3$ and $27n^3 - 2n + 1 = 5b^3$ for some positive integers $a$ and $b$.
From the second equation we conclude that $n$ cannot be divisible by $5$. If we have $n \equiv \pm1 \pmod 5$, then $27n^3 - 2n + 1 \equiv 1 \pmod 5$. If we have $n \equiv 2 \pmod 5$, then $27n^3 - 2n + 1 \equiv 3 \pmod 5$, and if $n \equiv 3 \pmod 5$, then $27n^3 - 2n + 1 \equiv 4 \pmod 5$. In all cases the left-hand side of the equation cannot be divisible by $5$.
Hence the equation has no solution.