Let $ n, m$ be positive integers of different parity, and $ n > m$.  Find all integers $ x$ such that $ \frac {x^{2^n} \minus{} 1}{x^{2^m} \minus{} 1}$ is a perfect square.
1. **Expression Simplification**:
   Given the expression:
   \[
   \frac{x^{2^n} - 1}{x^{2^m} - 1}
   \]
   We can factorize the numerator using the difference of powers:
   \[
   x^{2^n} - 1 = (x^{2^m} - 1)(x^{2^m} + 1)(x^{2^{m+1}} + 1) \cdots (x^{2^{n-1}} + 1)
   \]
   Therefore, the given expression simplifies to:
   \[
   \frac{x^{2^n} - 1}{x^{2^m} - 1} = (x^{2^m} + 1)(x^{2^{m+1}} + 1) \cdots (x^{2^{n-1}} + 1)
   \]

2. **Parity Analysis**:
   Since \( n \) and \( m \) have different parity and \( n > m \), the number of terms in the product \((x^{2^m} + 1)(x^{2^{m+1}} + 1) \cdots (x^{2^{n-1}} + 1)\) is odd.

3. **Odd \( x \)**:
   If \( x \) is odd, each term \( x^{2^k} + 1 \) (for \( k \geq m \)) is even because \( x^{2^k} \) is odd and adding 1 makes it even. Since there are an odd number of such terms, the product is divisible by \( 2^l \) where \( l \) is odd. This implies the product cannot be a perfect square because a perfect square must have an even power of 2 in its prime factorization.

4. **Even \( x \)**:
   If \( x \) is even, each term \( x^{2^k} + 1 \) is odd because \( x^{2^k} \) is even and adding 1 makes it odd. Each term is relatively prime to each other because they are of the form \( x^{2^k} + 1 \) and \( x^{2^j} + 1 \) for different \( k \) and \( j \). For the product to be a perfect square, each term must itself be a perfect square.

5. **Perfect Square Condition**:
   Consider \( x^{2^m} + 1 \). For this to be a perfect square, let:
   \[
   x^{2^m} + 1 = k^2
   \]
   for some integer \( k \). This implies:
   \[
   x^{2^m} = k^2 - 1 = (k-1)(k+1)
   \]
   Since \( x \) is even, \( x^{2^m} \) is a power of 2. The only way \( (k-1)(k+1) \) can be a power of 2 is if \( k-1 = 1 \) and \( k+1 = 3 \), giving \( k = 2 \). This implies:
   \[
   x^{2^m} = 2 \cdot 2 = 4 \implies x = 2^{1/2^m}
   \]
   However, \( x \) must be an integer, and the only integer solution is \( x = 0 \).

6. **Conclusion**:
   Therefore, the only integer \( x \) that satisfies the given condition is \( x = 0 \).

The final answer is \( \boxed{ x = 0 } \).