Example 21 (2006 Bulgarian National Mathematical Olympiad) Consider the set $A=\{1,2,3,4, \cdots, 2^n\}$, where $n \geqslant 2$. If the sum of two elements in set $A$ is a power of 2, then one of them must belong to the subset $B$. Find the number of subsets $B$ of set $A$.
Let $B$ be a subset that satisfies the given property. Since $1+3=2^{2}$, either 1 or 3 must belong to $B$.
If $1 \in B$, then $3 \notin B$.
We will prove by mathematical induction:
For any integer $t\left(0 \leqslant t2 \times 2^{l}=2^{l+1}\right)$.
Thus, $0<2^{\prime+1}-(4 s+5)<(4 s+5)$.
Let $x=4 s+5, y=2^{l+1}-(4 s+5)$, then
$x+y=2^{l+1}$.
Since $y$ is an integer of the form $4 m+3$, hence
$y \notin B, 4(s+1)+1 \in B$.
Similarly, $4(s+1)+3 \notin B$.
If $1 \notin B$, then $3 \in B$.
Similarly, integers of the form $4 t+1$ do not belong to $B$, and integers of the form $4 t+3$ belong to $B$.
Therefore, the odd numbers in subset $B$ are either all integers of the form $4 t+1$ or all integers of the form $4 t+3$.
Let $x=2^{p} x_{0}, y=2^{q} y_{0}$, where $x_{0}, y_{0}$ are odd numbers, and $p, q$ are positive integers.
If $2^{p} x_{0}+2^{q} y_{0}=2^{k}, p \neq q$, without loss of generality, assume $p<q$, then $x_{0}+2^{q-p} y_{0}=2^{k-p}$ is impossible, so, $p=q$. Therefore, the sum of elements taken from different sets $A_{i}=\left\{2^{i} a \mid a\right.$ is an odd number $\}(i=1,2, \cdots, n)$ is not a power of 2.

For any $A_{i}$, dividing by $2^{i}$ and from the above discussion, all integers of the form $2^{i}(4 t+1)$ are in $B$ or all integers of the form $2^{i}(4 t+3)$ are in $B$.
Therefore, there exist $2^{n+1}$ subsets $B$ that satisfy the given requirements.