$n=4,5$ has no solution $mod 16, mod 11$ respectively $(n,a,b,c)=(1,1,1,1),(2,1,1,0)$ so it suffices $n=3,6$

If $n=1$ we take $(a,b,c)=(1,1,1)$ If $n=2$ we take $(a,b,c)=(1,1,0)$ If $n=3$ we take $(a,b,c)=(1,1,-1)$ If $n=4$ $\Longrightarrow$ $a^4+b^4\equiv c^4+4\pmod 4$, since $x^{4}\equiv 0,1\pmod 4$ we get $a\equiv b\equiv c\equiv 0\pmod 2$ $\Longrightarrow$ $a^4+b^4\equiv 0\pmod 8\equiv c^4+4 \equiv 4\pmod 8$ which is a contradiction. If $n=5$ $\Longrightarrow$ $a^5+b^5\equiv c^5+5\pmod {11}$, since $x^{5}\equiv -1,0,1\pmod {11}$ we get a contradiction. If $n=6$ $\Longrightarrow$ $a^6+b^6\equiv c^6+6\pmod 8$, since $x^6\equiv 0,1\pmod 8$ we get a contradiction.

Note that it is easy to find solutions for $n=1,2,3$. For $n=4$ use $\pmod{4}$ to show that at least one of the $a$ and $b$ is divisible by 4 and then take $\pmod{8}$ to show no solutions. For $n=5$ take $\pmod{11}$. For $n=6$ you can also use $\pmod{13}$ instead of $\pmod{8}$.