Integers $n < 2$ obviously do not satisfy $2^n - n^3 = 4! = 24$. For $2 \le n \le 9$ we easily check that they satisfy $2^n < n^3$, hence $2^n - n^3 \ne 24$.
$$2^2 < 3^2 \qquad 2^4 < 2^6 = 4^3 \qquad 2^6 < 2^3 \cdot 3^3 = 6^3 \qquad 2^9 = 8^3 < 9^3$$
$$2^3 < 3^3 \qquad 2^5 < 2^6 = 4^3 < 5^3 \qquad 2^8 < 2^9 = 8^3 \qquad \text{and}$$
$$2^7 = 8 \cdot 4 \cdot 4 < 8 \cdot 6 \cdot 6 = 48 \cdot 6 < 49 \cdot 7 = 7^3.$$

Because $2^{10} = 1024 = 10^3 + 24$, $n = 10$ satisfies the condition $2^n - n^3 = 24$. For $n \ge 10$ we show that $2^{n+1} - (n+1)^3 > 2^n - n^3 > 0$. First note that
$$2^{n+1} - (n+1)^3 = 2 \cdot 2^n - n^3 - 3n^2 - 3n - 1 \\ > 2 \cdot 2^n - 2 \cdot n^3 = 2(2^n - n^3)$$
because $3n^2 + 3n + 1 = n^2(3 + \frac{3}{n} + \frac{1}{n^2}) < n^3$ for $n \ge 6$. Since $2^{10} - 10^3 = 24 > 0$ it follows now by induction that $2^n - n^3 > 0$ for $n \ge 10$ and so $2^{n+1} - (n+1)^3 > 2(2^n - n^3) > 2^n - n^3$ as claimed.
We have shown that $n = 10$ is the only integer that satisfies $2^n - n^3 = 24$. The only possible value for $2^n + n^3$ therefore is $2^{10} + 10^3 = 2024$.