Suppose $n\geq2$ satisfied the condition ($n=1$ clearly fails). Cubes are $0,\pm1\pmod9$, so $2^n\equiv\pm1\pmod9\implies3\mid n$. But if $n=3k$, then $(3^k)^3<2^{3k}+3^{3k}<\left(3^k+1\right)^3$, contradiction.
Mx.22 wrote:
Can someone give a solution whith (mod4).
Imposible, you're letting out the properties of $2$ because $2^n\equiv 0\mod 4$ for any $n\ge 2$ and since $3^n\equiv (1,3)\mod 4$ this isn't give particular properties for perfect cubes.