Find all pairs $(m,n)$ of positive integers numbers with $m>1$ such that:
For any positive integer $b \le m$ that is not coprime with $m$, its posible choose positive integers $a_1, a_2, \cdots, a_n$ all coprimes with $m$ such that:
$$m+a_1b+a_2b^2+\cdots+a_nb^n$$
Is a perfect power.
Note: A perfect power is a positive integer represented by $a^k$, where $a$ and $k$ are positive integers with $k>1$
1. **Lemma: \( m \) must be squarefree.**

   Assume not and take \( m = p^2k \) for some prime \( p \). Now, taking \( b = p \), we get that:
   \[
   A = p^2k + a_1p + a_2p^2 + \cdots + a_np^n
   \]
   must be a perfect power. We can see that \( v_p(p^2k) \geq 2 \) and \( v_p(a_ip^i) \geq 2 \) for all \( i \in \{2, \ldots, n\} \), meanwhile \( v_p(a_1p) = 1 \) because \( \gcd(a_1, p) = 1 \). This means that \( v_p(A) = 1 \) and therefore \( A \) cannot be a perfect power. This contradiction implies that \( m \) must be squarefree. \(\blacksquare\)

2. **Claim: \( k = 1 \), i.e., \( m \) must be prime.**

   Assume for the sake of contradiction that \( k \geq 2 \), and assume that \( p_1 < p_2 < \cdots < p_k \). Then \( p_1^2 \leq p_1p_2 \cdots p_k \), so we may take \( b = p_1^2 \). Let \( p_1 = p \) (for simplicity). Then:
   \[
   A = pp_2 \cdots p_k + a_1p^2 + a_2p^4 + \cdots + a_np^{2n}
   \]
   must be a perfect power. We can see that \( v_p(pp_2 \cdots p_k) = 1 \), meanwhile, \( v_p(a_ip^{2i}) \geq 2 \) for all \( i \in \{1, \ldots, n\} \). Then \( v_p(A) = 1 \) meaning that \( A \) cannot be a perfect power. This contradiction implies that \( m \) must be prime. \(\blacksquare\)

3. **Construction for Primes:**

   Write \( m = p \) for some prime \( p \). We get that the only possible value of \( b \) is \( p \) because \( \gcd(p, b) \neq 1 \) and \( b \leq p \). Set \( a_i = p-1 \) for all \( i \in \{1, \ldots, n\} \). Then:
   \[
   A = p + (p-1)p + (p-1)p^2 + \cdots + (p-1)p^n = p + p^2 - p + p^3 - p^2 + \cdots + p^{n+1} - p^n = p^{n+1}
   \]
   This is clearly a perfect power.

The final answer is \( \boxed{ (m, n) } \) where \( m \) is a prime number, \( n \in \mathbb{N} \).