Erdős [Er69] gave a simple proof that $F(n) \leq \pi(n)+n^{2/3}$: we define a graph with vertex set the union of those integers in $[1,n^{2/3}]$ with all primes $p\in (n^{2/3},n]$. We have an edge $u\sim v$ if and only if $uv\in A$. It is easy to see that every $m\leq n$ can be written as $uv$ where $u\leq n^{2/3}$ and $v$ is either prime or $\leq n^{2/3}$, and hence there are $\geq \lvert A\rvert$ many edges. This graph contains no path of length $3$ and hence must be a tree and have fewer edges than vertices, and we are done. This can be improved to give the upper bound mentioned by using a subset of integers in $[1,n^{2/3}]$.
More generally, one can ask for such an asymptotic for the size of sets such that no $a\in A$ divides the product of $r$ distinct other elements of $A$, with the exponent $2/3$ replaced by $\frac{2}{r+1}$.
See also [425].