Let $\mathbb{N}$ denote the set of all positive integers. An ordered pair $(a;b)$ of numbers $a,b\in\mathbb{N}$ is called [i]interesting[/i], if for any $n\in\mathbb{N}$ there exists $k\in\mathbb{N}$ such that the number $a^k+b$ is divisible by $2^n$. Find all [i]interesting[/i] ordered pairs of numbers.
1. **Define the valuation function**: For any positive integer \( n \), let \( v(n) \) be the largest integer such that \( 2^{v(n)} \mid n \).

2. **Lemma**: If \( b \equiv 1 \pmod{2^{v(a-1)}} \), then there exists \( k \in \mathbb{N} \) such that \( a^k \equiv b \pmod{2^n} \) for any \( n \in \mathbb{N} \).

3. **Proof of Lemma**:
    - Let \( v(a-1) = p \), \( a = 2^p l + 1 \) where \( l \) is odd, and \( b = 2^p m + 1 \).
    - For \( n \le p \), the statement is obvious (just take \( k = 1 \)).
    - Suppose \( n > p \). Note that:
      \[
      a^{2^{n-p}} - 1 = \sum_{i=1}^{2^{n-p}} (2^p l)^i \binom{2^{n-p}}{i}
      \]
    - Note that \( v((2^p l)^i \binom{2^{n-p}}{i}) \ge pi + n - p - v(i) > n \). So \( a^{2^{n-p}} \equiv 1 \pmod{2^n} \).
    - If \( a, a^2, \ldots, a^{2^{n-p}} \) have distinct values modulo \( 2^n \), then they take all values of the form \( 2^p m + 1 \), so \( a^k \equiv b \pmod{2^n} \) for some \( k \).
    - Otherwise, the order of \( a \) modulo \( 2^n \) is less than \( 2^{n-p} \), thus we must have \( a^{2^{n-p-1}} \equiv 1 \pmod{2^n} \).
    - But we have:
      \[
      a^{2^{n-p-1}} - 1 = \sum_{i=1}^{2^{n-p-1}} (2^p l)^i \binom{2^{n-p-1}}{i}
      \]
    - For \( i \ge 2 \), we have \( v((2^p l)^i \binom{2^{n-p-1}}{i}) = pi + n - p - 1 - v(i) \ge n \), and for \( i = 1 \), we have \( v((2^p l)^i \binom{2^{n-p-1}}{i}) = v(2^p l 2^{n-p-1}) = n - 1 \). So \( 2^n \nmid a^{2^{n-p-1}} - 1 \), a contradiction. Our lemma is proved.

4. **Parity Consideration**:
    - Since \( a^k + b \) is even, then \( a \) and \( b \) have the same parity.
    - Suppose they are both even. For \( n > v(b) \), we have \( a^k \equiv -b \pmod{2^n} \). So we must have \( k v(a) = v(b) \), or \( k = \frac{v(a)}{v(b)} \). Thus \( a^{\frac{v(a)}{v(b)}} + b \) is divisible by \( 2^n \) for all \( n > v(b) \), which is impossible. Therefore \( a \) and \( b \) are both odd.

5. **Divisibility Consideration**:
    - For \( n = v(a^2 - 1) \), we have \( a^k \equiv -b \pmod{2^n} \). Note that either \( a^2 - 1 \mid a^k - 1 \) or \( a^2 - 1 \mid a^k - a \). Thus \( -b - 1 \) or \( -b - a \) is divisible by \( 2^{v(a^2 - 1)} \), equivalently, \( 2^{v(a^2 - 1)} \) divides \( b + 1 \) or \( a + b \).

6. **Case Analysis**:
    - **Case (i)**: \( a \equiv 1 \pmod{4} \)
        - Let \( a = 2^k l + 1 \) where \( k \ge 2 \) and \( l \) is odd. Then \( a^2 - 1 = 4^k l^2 + 2^{k+1} l \), so \( v(a^2 - 1) = k + 1 \).
        - So \( b = 2^{k+1} m - 1 \) or \( b = 2^{k+1} m - 2^k l - 1 = 2^k (2m - l) - 1 = 2^k p - 1 \). Either way, we have \( b = 2^k q - 1 \) for some positive integer \( q \).
        - From our lemma, we know that there exists \( x \in \mathbb{N} \) such that \( a^x \equiv -b \pmod{2^n} \), so \( (2^k l + 1, 2^k q - 1) \) is an interesting pair for any \( k \ge 2 \) and odd \( l \).

    - **Case (ii)**: \( a \equiv -1 \pmod{4} \)
        - Let \( a = 2^k l - 1 \) where \( k \ge 2 \) and \( l \) is odd. Then \( a^2 - 1 = 4^k l^2 - 2^{k+1} l \), so \( v(a^2 - 1) = k + 1 \).
        - Then \( b = 2^{k+1} m - 1 \) or \( b = 2^{k+1} m - 2^k l + 1 = 2^k q + 1 \) where \( q \) is odd.

        - **Subcase (ii.i)**: \( b = 2^{k+1} m - 1 \)
            - Note that \( v(a^2 - 1) = v(4^k l^2 - 2^{k+1} l) = k + 1 \). Since \( -b \equiv 1 \pmod{2^{k+1}} \), then it follows from our lemma that there exists \( x \) such that \( (a^2)^x \equiv -b \pmod{2^n} \). So \( (2^k l - 1, 2^{k+1} m - 1) \) is interesting for \( k \ge 2 \) and odd \( l \).

        - **Subcase (ii.ii)**: \( b = 2^k q + 1 \)
            - Consider a sufficiently large \( n \). It is easy to see that the inverse of \( a \) modulo \( 2^n \) is of the form \( a^* \equiv 2^k r - 1 \) for odd \( r \). Since \( -b a^* \equiv 1 \pmod{2^{k+1}} \), from our lemma we know that there exists \( x \) such that \( (a^2)^x \equiv -b a^* \pmod{2^n} \), which gives \( a^{2x+1} \equiv -b \pmod{2^n} \). Therefore \( (2^k l - 1, 2^k q + 1) \) is interesting for \( k \ge 2 \) and odd \( l, q \).

The final answer is \( \boxed{ (2^k l + 1, 2^k q - 1) } \) and \((2^k l - 1, 2^k q + 1)\) for \( k \ge 2 \) and odd \( l, q \).