A $\frac{1}{3}$ Conjecture?

As in @Winther's comment, I will focus on the asymptotic density of perfect powers.

To be specific, let the number of perfect powers which does not exceed $n$ be $\alpha(n)$. If we can prove $\frac{\alpha(2m)}{m}<\frac{1}{3}$ for all $m>N$, then we need to check all $6<n\le N$ manually since we can prove $$\frac{\alpha(2m)}{m}\ge{1 \above 1.5 pt m^2}\sum_{i=1}^m \sum_{j=1}^ma_{ij}$$

The proof of above inequality is simple. Since there are at most $m$ solutions to $i+j=k, 1\le i,j \le m$, one perfect power "contributes" at most $m$ to the sum of $a_{ij}$s. It follows ${1/ m^2}\sum_{i=1}^m \sum_{j=1}^ma_{ij} \le (1/m^2) m\alpha(2m)=\alpha(2m)/m$.

There are less than $\sqrt{2n}$ square numbers and $\sqrt[3]{2n}$ cubic numbers which is not $1$ and does not exceed $2n$, and $\sqrt[5]{2n}$ 5-power numbers (note that every 4-power numbers are also squares). Also, since $2^{\log_2{2n}}=2n$, the category of power numbers are less than $\log_2{2n}$. Therefore, the number of perfect powers which is not $1$ and does not exceed $2n$ is at most:$$\sqrt{2n}+\sqrt[3]{2n}+(\log_2{2n}-4)\sqrt[5]{2n}$$ And one can check that this does not exceed $\frac{n}{3}$ if $n \ge 151$. Now one can manually check the ratio does not exceed $\frac{1}{3}$ for small $n$s except $1$ to $30$ and conclude the proof.