Every minimal blocking set in $PG(2,q)$ has this property, i.e, point sets which intersect every line, do not contain any line, and are minimal with respect to inclusion. This is because if there is a point $x$ in such a minimal blocking set through which there is no tangent, then you can just remove this point $x$ to get a proper subset of the blocking set which is also a blocking set.

You are welcome. But if we use the method that you have described for a general $q$, then shouldn't the bound be $(q^2 - 1)^{\log_2^n} = n^{\log_2^{q^2 - 1}}$, when $n$ is a power of $2$? This is because in your base step you will have $F(2) = q^2 - 1$. And then this upper bound will be worse than the quadratic upper bound we have. For example, for $q = 3$ it gives an upper bound of $O(n^3)$, while the bounds I have described are $O(n^2)$.