Let $ n > 1, n \in \mathbb{Z}$ and $ B \equal{}\{1,2,\ldots, 2^n\}.$ A subset $ A$ of $ B$ is called weird if it contains exactly one of the distinct elements $ x,y \in B$ such that the sum of $ x$ and $ y$ is a power of two. How many weird subsets does $ B$ have?
1. **Define the problem and notation:**
   We are given a set \( B = \{1, 2, \ldots, 2^n\} \) and need to find the number of subsets \( A \subseteq B \) that are "weird." A subset \( A \) is called weird if it contains exactly one of the distinct elements \( x, y \in B \) such that \( x + y \) is a power of two.

2. **Establish the recurrence relation:**
   Let \( f(n) \) denote the number of weird subsets of \( B \) when \( B = \{1, 2, \ldots, 2^n\} \). We aim to show that \( f(n) = 2^{n+1} \).

3. **Base case:**
   For \( n = 2 \), \( B = \{1, 2, 3, 4\} \). The pairs \((x, y)\) such that \( x + y \) is a power of two are:
   - \( 1 + 3 = 4 \)
   - \( 2 + 2 = 4 \) (not distinct)
   - \( 4 + 0 = 4 \) (not in the set)
   
   The valid pairs are \((1, 3)\). The weird subsets are:
   - \(\{1\}\)
   - \(\{2\}\)
   - \(\{3\}\)
   - \(\{4\}\)
   - \(\{1, 2\}\)
   - \(\{1, 4\}\)
   - \(\{2, 3\}\)
   - \(\{3, 4\}\)
   
   Thus, \( f(2) = 8 = 2^3 \).

4. **Inductive step:**
   Assume \( f(n) = 2^{n+1} \) holds for some \( n \geq 2 \). We need to show that \( f(n+1) = 2^{(n+1)+1} \).

   Consider the set \( B' = \{1, 2, \ldots, 2^{n+1}\} \). We can split \( B' \) into two parts:
   - \( B_1 = \{1, 2, \ldots, 2^n\} \)
   - \( B_2 = \{2^n + 1, 2^n + 2, \ldots, 2^{n+1}\} \)

   Each weird subset of \( B' \) can be constructed by taking a weird subset of \( B_1 \) and either including or excluding \( 2^{n+1} \). This gives us two choices for each weird subset of \( B_1 \).

   Therefore, the number of weird subsets of \( B' \) is:
   \[
   f(n+1) = 2 \cdot f(n)
   \]

5. **Verify the recurrence relation:**
   Given \( f(n) = 2^{n+1} \), we have:
   \[
   f(n+1) = 2 \cdot 2^{n+1} = 2^{(n+1)+1}
   \]

6. **Conclusion:**
   By induction, we have shown that \( f(n) = 2^{n+1} \) for all \( n \geq 2 \).

\(\blacksquare\)

The final answer is \( \boxed{ 2^{n+1} } \)