Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. However, this isn't quite good enough. The number of $2$ groups of order $n = 2^m$ is $n^{B \log n}$ where              

$$B = \frac{2}{27 \log 2}.$$

You would win if $B > A$, but Lubotzky doesn't compute $A$, and a quick look at the argument suggests that the bound is not going to be in your favor. Looking at papers which cite Lubotzky's paper I do not quickly see any better bounds in the literature, so you may be out of luck.

Here is Lubotzky's paper: https://core.ac.uk/download/pdf/82373534.pdf