Determine all positive integers $m$ satisfying the condition that there exists a unique positive integer $n$ such that there exists a rectangle which can be decomposed into $n$ congruent squares and can also be decomposed into $m+n$ congruent squares.
1. **Restate the problem in mathematical terms:**
   We need to find all positive integers \( m \) such that there exists a unique positive integer \( n \) for which a rectangle can be decomposed into \( n \) congruent squares and also into \( m+n \) congruent squares.

2. **Reformulate the problem:**
   The problem can be rephrased as finding all \( m \) such that there exists exactly one positive divisor \( d \) of \( m \) which can be expressed as \( a^2 - b^2 \) where \( \gcd(a, b) = 1 \).

3. **Analyze the condition \( a^2 - b^2 \):**
   Note that \( a^2 - b^2 = (a+b)(a-b) \). For \( d \) to be expressed as \( a^2 - b^2 \), \( d \) must be the product of two coprime factors. 

4. **Consider the number of odd divisors:**
   If \( m \) has two or more odd divisors, it cannot satisfy the condition because every positive integer \( 2k+1 = (k+1)^2 - k^2 \) and \( \gcd(k+1, k) = 1 \). Thus, \( m \) must have exactly one odd divisor.

5. **Form of \( m \):**
   Therefore, \( m \) must be of the form \( 2^k p \) where \( p \) is an odd prime. 

6. **Check the constraints on \( k \):**
   Since \( 8 = 3^2 - 1 \), \( k \leq 2 \). This is because if \( k \geq 3 \), \( m \) would have more than one odd divisor.

7. **Verify the forms of \( m \):**
   - For \( k = 0 \), \( m = p \).
   - For \( k = 1 \), \( m = 2p \).
   - For \( k = 2 \), \( m = 4p \).

8. **Conclusion:**
   The integers \( m \) that satisfy the given condition are \( p \), \( 2p \), and \( 4p \) where \( p \) is an odd prime.

The final answer is \( \boxed{ p } \), \( 2p \), and \( 4p \) where \( p \) is an odd prime.