In "A Classical Introduction to Modern Number Theory" by Ireland and Rosen, a very close bound is obtained by completely elementary means.
Assum $p_n \le x < p_{n+1}$, where $p_i$ denotes the $i$'th prime. By definition, $\pi(x) = n$.
Considers the numbers up to $x$: $\{1,2,\cdots, x \}$. They are divisible only by primes among $\{ p_1, p_2, \cdots, p_{n} \}$. If you decompose those numbers into a square part and square-free part, i.e. write $a = rs^2$ where $r$ is a product of distinct primes and $s$ is a square, we see that there are 2 constraints on $r,s$:
Since $r$ is square-free, it must be a product of distinct primes among the first $n$ primes. There are only $2^{n}$ options for $r$.
Since $s^2 \le a$, we must have $s \le \sqrt{a} \le \sqrt{x}$, so there are at most $\sqrt{x}$ options for $s$.
All in all, there are at most $2^n \times \sqrt{x}$ options for those numbers between $1$ and $x$: $ x \le 2^{n} \sqrt{x} = 2^{\pi(x)} \sqrt{x}$
This shows that $\pi(x) \ge \log_{2} \sqrt{x} = \frac{\log_{2} x}{2}$. $\blacksquare$
Similarly, if we decompose those numbers into an $n$'th-power and an $n$-powerfree number, we find: $\pi(x) \ge \frac{\ln x (1-\frac{1}{n}) }{\ln n} $ But $n=2$ already gives the best bound. I will think later about improving this to $\pi(x) \ge \log_2 x$ - I believe it is possible with a similar method.