Paul Erdős showed a simple estimate for $\pi(x) \ge \frac{1}{2}\log_2 x$; is it possible to tweak his argument to improve the estimate? Paul Erdős gave a simple argument to show that $\pi(x) \ge \dfrac{1}{2}\log_2 x$.
Is it possible to tweak the argument and get a better estimate?  I am wondering how good an estimate for $\pi(x)$ can be achieved using a variation of his reasoning.
I explored the possibility of changing $m^2$ to $m^3$ so that we have for any $y \le x$, we have:
$y = p_1^{e_1}p_2^{e_2}\ldots p_n^{e_n}m^3$ and $e_i\in\left\{0,1,2\right\}$ and $m \in \mathbb{Z}$
But this gets us to:  $m \le \sqrt[3]{x}$ so that $3^n\times \sqrt[3]{x} \ge x$ and $\pi(x) \ge \dfrac{2}{3}\log_3 x$ which is weaker than $\pi(x) \ge \dfrac{1}{2}\log_2 x$
Has anyone thought of other creative tweaks that can improve the result? 

The same idea also gives $p_n\leq 4^n$ and a rather trivial improvement of that is:
$n\in\Bbb Z_+,\;1\leq x\leq p_n\implies x=p_1^{e_1}\cdots p_n^{e_n}\cdot m^2,\;e_i=0,1\wedge\;m^2<p_n$.
If $e_n=1$ then $0=e_1=\cdots=e_{n-1}$ and if $e_n=0$ there is at most $2^{n-1}$ ways to chose $x$. Similar to Erdős proof $p_n\leq(2^{n-1}+1)\sqrt p_n$ which gives 
$p_n\leq (2^{n-1}+1)^2$.
A challenge would be to prove $p_n\leq 2^{n-1}+1$ without using the prime number theorem.
