Find all pairs $(a,b)$ of positive integers, such that for [b]every[/b] $n$ positive integer, the equality $a^n+b^n=c_n^{n+1}$ is true, for some $c_n$ positive integer. 
1. **Claim:** We cannot have an odd prime \( p \) and an odd positive integer \( n \) such that \( a \) and \( b \) are not divisible by \( p \) and \( p \mid a^n + b^n \).

   **Proof:**
   Assume for contradiction that the claim is false. Then let \( p \) and \( n \) be as above. Since \( n \) is odd and \( p \) does not divide \( a \) and \( b \), we must have by the Lifting The Exponent (LTE) lemma that for all odd positive integers \( k \):
   \[
   v_p(a^{nk} + b^{nk}) = v_p(a^n + b^n) + v_p(k)
   \]
   On the other hand, by the problem's condition:
   \[
   nk + 1 \mid v_p(a^{nk} + b^{nk})
   \]
   Combining the two relations, we get an absurdity since we can make \( nk + 1 \) arbitrarily large while \( v_p(a^n + b^n) + v_p(k) \) is nonzero and can be small (for instance, take \( k \) sufficiently large with \( p \nmid k \)). This contradiction proves the claim. \(\blacksquare\)

2. **Lemma:** If \( a \) and \( b \) are distinct positive integers, then there exists an odd integer \( n \) and an odd prime \( p \) such that \( p \nmid a, b \) and \( p \mid a^n + b^n \).

   **Proof:**
   First, write \( d = \gcd(a, b) \) and then let \( x = \frac{a}{d} \) and \( y = \frac{b}{d} \). Take \( q \) to be a sufficiently large odd prime, more precisely, take \( q \) so that \( q > a + b \) and \( q > r \) for all \( r \mid d \). Then take \( n = q \) and \( p \) as an odd prime dividing:
   \[
   \frac{x^q + y^q}{x + y}
   \]
   To see that such \( p \) exists, first note that since \( a \ne b \) by assumption, we cannot have \( x = y = 1 \) and hence \(\frac{x^q + y^q}{x + y} > 1\). Now to show that \( p \) is odd, note that since \( \gcd(x, y) = 1 \), we must have that either \( x, y \) are both odd or have different parities. In the latter case, we would have that \(\frac{x^q + y^q}{x + y}\) is odd and we are done; in the former case, just note that:
   \[
   \frac{x^q + y^q}{x + y} = x^{q-1} - x^{q-2}y + \dots - xy^{q-2} + y^{q-1}
   \]
   is odd since it is the sum of an odd number of odd numbers. So indeed \( p \) exists and is odd.

   Now we have to show that \( p \) satisfies the other condition, i.e., \( p \nmid a, b \). First note that since \( x \) and \( y \) are coprime, we have \( p \nmid x, y \) and so \( p \mid a, b \iff p \mid d \). Now note that since \( q \) is odd, we have that:
   \[
   p \mid x^q + y^q \Rightarrow (-xy^{-1})^q \equiv 1 \pmod{p}
   \]
   where \( y^{-1} \) denotes the multiplicative inverse of \( y \) modulo \( p \). Then let \( k \) denote the order of \( -xy^{-1} \) modulo \( p \). By the above congruence, we have that \( k \mid q \) and since \( q \) is prime, it follows that \( k = q \) or \( k = 1 \). If the former case happens, then we have that \( p - 1 \ge k = q \) which implies that \( p \nmid d \) by the choice of \( q \) (we have selected \( q \) so that it is greater than all divisors of \( d \)), hence \( p \nmid a, b \) and we are done. In the latter case, we have that \( p \mid x + y \); hence by LTE we have:
   \[
   v_p\left(\frac{x^q + y^q}{x + y}\right) = v_p(q)
   \]
   Since \( p \) divides the above number by definition, we must have that \( v_p(q) \ge 1 \), i.e., \( p = q \). Then \( q \mid x + y \mid a + b \) which is an absurdity since we have chosen \( q > a + b \). So the lemma is proven. \(\blacksquare\)

3. From the **Claim** and the **Lemma**, it follows that we must have \( a = b \) and the condition becomes \( 2a^n \) is an \( n+1 \) power for all \( n \). We claim that \( a \) must be \( 2 \). Indeed, let \( l = v_2(a) \), then the problem's condition becomes \( n + 1 \mid 1 + nl \) for all positive integers \( n \). This is equivalent to \( n + 1 \mid l - 1 \) for all positive integers \( n \), which can only happen if \( l - 1 = 0 \) (since \( l \) is fixed and \( n \) can be arbitrarily large), i.e., \( l = 1 \).

4. Now let \( p \) be any odd prime, then the condition implies \( n + 1 \mid nv_p(a) \) for all positive integers \( n \); since \( n + 1 \) and \( n \) are coprime, this is equivalent to \( n + 1 \mid v_p(a) \) for all positive integers \( n \), which similarly implies \( v_p(a) = 0 \). Then we have that \( v_2(a) = 1 \) and \( v_p(a) = 0 \) for all odd primes \( p \), i.e., \( a = 2 \). Clearly, \( (2, 2) \) satisfies the problem's condition and so we are done.

The final answer is \( \boxed{ (2, 2) } \).