Saudi Arabian Mathematical Competitions

For any integer $a$ we have $a^{3} \equiv 0,1,8 \pmod{9}$, hence for any integers $x$ and $y$, we have
$$x^{3}+y^{3} \equiv 0,1,2,7,8 \quad (\bmod 9).$$
For $n \geq 6$, we have $n!+4 \equiv 4 \pmod{9}$, that is there are no solutions in this case.

If $n=1$, then the equation becomes $x^{3}+y^{3}=5$, which has no solutions because of the possible residues above.

If $n=2$, then we get the equation $x^{3}+y^{3}=6$, which has no solutions because of the possible residues above.

If $n=3$, then the equation becomes $x^{3}+y^{3}=10$.
For any integer $a$ we have $a^{3} \equiv 0, \pm 1 \pmod{7}$, hence for any integers $x$ and $y$, we have
$$x^{3}+y^{3} \equiv 0, \pm 1, \pm 2 \quad (\bmod 7).$$
The equation has no solutions because of the possible residues above.

If $n=4$, then the equation becomes $x^{3}+y^{3}=28$, with solution $(x, y)=(3,1)$.

If $n=5$, then we get the equation $x^{3}+y^{3}=124$, with solution $(x, y)=(5,-1)$.

The equation has solutions in integers if and only if $n=4$ and $n=5$.