What is the minimum number of people such that the probability for some of them to have a common birthday is $\frac{1}{2}$?
So looking at the problem as "balls and bins". We have $n = 365$ bins and $m$ balls (people).
There's a probability of $1/n$ for two balls to be on the same bin.
Hence, the expected number of couples to be on the same bin is $\frac{m(m-1)}{2n}$
How can I leverage this information to conclude that the desired $m$ is $23$?
You can't conclude any exact result from this, since the expected number includes contributions from cases with more than one collision. However, when the probability of at least one collision is $\frac12$, the probability of more than one collision is still rather low, so you can get an estimate by taking the expected number as an approximation of the probability of at least one collision. Then
$$ \frac{m(m-1)}{2n}\ge\frac12 $$
yields
$$ m^2-m-n\ge0 $$
and thus
$$ m\ge\frac12+\sqrt{n+\frac14}\;, $$
which for $n=365$ yields $m\ge20$, not a bad estimate.