The answer is: All $(m,n)$ where $m$ is a prime number, $n \in \mathbb{N}$.

Lemma: $m$ must be squarefree. Assume not and take $m = p^2k$ for some $p$ prime. Now, taking $b = p$, we get that: $$A = p^2k+a_1p+a_2p^2+...+a_np^n$$must be a perfect power. We can see that $v_p(p^2k) \geq 2, v_p(a_ip^i) \geq 2$ for all $i \in \{2,..,n\}$, meanwhile $v_p(a_1p) =1$ because $gcd(a_1,p) =1$ meaning that $v_p(A) =1$ and therefore $A$ can't be a perfect power. $\square$ By the Lemma, write $m = p_1p_2...p_k$ for some $k \in \mathbb{N}$. Claim: $k=1$, i.e, $m$ must be prime. FTSOC assume that $k \geq 2$, and assume that $p_1 < p_2 < ... <p_k$, then $p_1^2 \le p_1p_2....p_k$, so we may take $b=p_1^2$. Let $p_1 =p$ (for latexing purposes) Then $$A = pp_2...p_k + a_1p^2 + a_2p^4+...+a_np^{2n}$$must be a perfect power. We can see that $v_p(pp_2....p_k) = 1$, meanwhile, $v_p(a_ip^{2i}) \geq 2$ for all $i \in \{1,...,n \}$. Then $v_p(A) = 1$ meaning that $A$ can't be a perfect power. $\square$ 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 $i \in \{1,...,n \}$ Then $$A = p+(p-1)p+(p-1)p^2+...+(p-1)p^n = p + p^2 - p + p^3 - p^2 + ... + p^{n+1} - p^n = p^{n+1}$$This is clearly a perfect power.

I think this was relatively good in terms of difficulty for a TST P1, it consists of three relatively natural steps. I would be happy to see other solutions