Bump Any Idea?

Ideas: Let $P(d)$ be the given condition. Let $m=\prod_{i=1}^{\omega(m)} p_i^{\alpha_i}$. If $\nu_p(m)>1$ for some $p\mid m$ so $\nu_q(m)=1$ for all other $q\neq p$ $\textit{Proof.}$ Suppose $q^2\mid m$ so $P(pq^2)\implies p\left(\frac{m}{p}+\sum_{i=1}^{2019} a_i(pq^2)^{i-1}p^{-1}\right)=M$ is a perfect power, moreover notice $gcd(p,\frac{m}{p}+\sum_{i=1}^{2019} a_i(pq^2)^{i-1}p^{-1})=\gcd\left(p,\frac{m}{p}+a_1q^2\right)=\gcd(p,a_1q^2)=1$ so $M$ is a perfect $1-th$ power (each prime has $\nu_p$ equals 1), moreover easy to see $q^2\mid M$ contradiction. So $m=q^k \prod_{i=1}^{\omega(m)-1} p_i$.

If $p^2|m$ choose $d=p$ and you will get $p|a_1$, contradiction, so $m$ is square free. If $m=p_1p_2...$, with $p_1$ being the smaller, choose $d=p_1^2$ and you will get contradiction again, so $m$ should be prime. If $m=p$, with $p$ prime, the only option for $d$ is $p$, and then you set $a_i=p-1$ and you get, \[ p+(p-1)p+(p-1)p^2+\dots (p-1)p^{2019}=p+p^2-p+p^3-p^2+\dots p^{2020}-p^{2019}=p^{2020} \] Then the answer is every prime $m$.

Huh? Sorry, if we have $n=p$ can we say $n$ is a $1-th$ power? Or a perfect power is strictly $n^k$ with $k\ge 2$?

Perfect power means $k\geq 2$ otherwise every number is a perfect power.