I think it is quite a nice problem for a JBMO... When $n=3$ Setting $k=l=m$ produces a square. When $n=4$ Note that $4^k+4^l+4^m=2^{2k}+2^{2l}+2^{2m}$ which is a square when for $(k,l,m)=(k,k,k+1)$ When $n=5$ Then $5^k+5^l+5^n=3 \pmod{4}$, hence it is never a square. When $n=6$ We have $6^k+6^l+6^n=8 \pmod{10}$, hence it is never a square. When $n=7$ For $(k,l,m)=(2k,2k,2k+1)$ we have infinitely many squares. Therefore: i): $n\in \{2,3,7 \} $ ii) $n\in \{5,6\} $