ii. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square of a natural number.
Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$

Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.

By verifying $n=2,4,6,8$ one by one, we find that $N$ is not a square number in any of these cases.
(2) When $n \geqslant 9$, $N=2^{8}\left(9+2^{n-8}\right)$.
For $N$ to be a square number, $9+2^{n-8}$ must be the square of an odd number.
Assume $9+2^{n-8}=(2 k+1)^{2}$. Hence,
$$
2^{n-10}=(k-1)(k+2) \text {. }
$$

Since $k-1$ and $k+2$ are one odd and one even, the left side of the equation is a power of 2, so it must be that $k-1=1$, i.e., $k=2$.
From $2^{n-10}=2^{2}$, we know that $n=12$ is the solution.