Find all pairs of natural numbers $a,b$ , with $a\ne b$ , such that $a+b$  and $ab+1$  are powers of $2$.
We need to find all pairs of natural numbers \(a, b\) with \(a \ne b\) such that \(a + b\) and \(ab + 1\) are powers of 2. Let's denote these powers of 2 as \(2^p\) and \(2^q\) respectively. Thus, we have:
\[ a + b = 2^p \]
\[ ab + 1 = 2^q \]

1. **Case 1: \(a = 1\)**

   If \(a = 1\), then:
   \[ 1 + b = 2^p \implies b = 2^p - 1 \]
   \[ ab + 1 = 1 \cdot b + 1 = b + 1 = 2^p \]
   Since \(b = 2^p - 1\), we have:
   \[ b + 1 = 2^p \]
   This is indeed a power of 2, so \((1, 2^p - 1)\) is a solution.

2. **Case 2: \(a > 1\)**

   Since \(a > 1\), we have:
   \[ (a - 1)(b - 1) \geq 1 \]
   \[ ab + 1 \geq a + b \]
   Since \(a + b = 2^p\) and \(ab + 1 = 2^q\), we have:
   \[ 2^q \geq 2^p \implies q \geq p \]
   Since \(a + b\) divides \(ab + 1\), we have:
   \[ a + b \mid ab + 1 \]
   \[ a + b \mid ab + 1 - (a + b)a \]
   \[ a + b \mid -a^2 + 1 \]
   \[ a + b \mid a^2 - 1 \]
   Since \(a + b = 2^p\), we have:
   \[ 2^p \mid (a - 1)(a + 1) \]
   Since \(2^p\) is a power of 2, \(a\) must be odd. Therefore, \(a\) can be written as:
   \[ a = 2^{p-1}t \pm 1 \]
   where \(t\) is an integer. Since \(a < 2^p\), we have \(t = 1\). Thus:
   \[ a = 2^{p-1} \pm 1 \]

   - If \(a = 2^{p-1} + 1\), then:
     \[ b = 2^p - a = 2^p - (2^{p-1} + 1) = 2^{p-1} - 1 \]
     This implies \(a > b\), which contradicts our assumption that \(a < b\).

   - If \(a = 2^{p-1} - 1\), then:
     \[ b = 2^p - a = 2^p - (2^{p-1} - 1) = 2^{p-1} + 1 \]
     We need to check if \((2^{p-1} - 1)(2^{p-1} + 1) + 1\) is a power of 2:
     \[ (2^{p-1} - 1)(2^{p-1} + 1) + 1 = 2^{2(p-1)} - 1 + 1 = 2^{2(p-1)} \]
     This is indeed a power of 2.

Therefore, the pairs \((1, 2^p - 1)\) and \((2^{p-1} - 1, 2^{p-1} + 1)\) are the only solutions.

The final answer is \( \boxed{ (1, 2^k - 1) } \) and \((2^k - 1, 2^k + 1)\)