Taking $n = 10$ gives $2^n + n^3 = 1024 + 1000 = 2024 = 88 \cdot 23$, so $88$ divides $2^{10} + 10^3$.

We consider divisibility by $11$ and by $8$ separately. By Fermat's little theorem, we have $2^{10} \equiv 1 \pmod{11}$, whereas $10^3 \equiv (-1)^3 = -1 \pmod{11}$. In summary $2^{10} + 10^3 \equiv 1 - 1 = 0 \pmod{11}$, meaning that $11$ divides $2^{10} + 10^3$. As $8$ divides both $2^{10}$ and $10^3$, it divides the sum $2^{10} + 10^3$. Thus $88$ divides $2^{10} + 10^3$, as desired.