There is another approximation based on a Poisson distribution.
I was using this method in the 1990s (I have found my implementation in Java), but unfortunately do not remember where I first read about it.
It is also described in https://stats.stackexchange.com/questions/1308/extending-the-birthday-paradox-to-more-than-2-people.

Let $b$ be the number of days in the year, $n$ the number of people in the room,
and suppose you want to know the probability that at least $k + 1$ people share a birthday. (In the original question above, $b = 365$ and $k = 2.$)

Let $X_i$ be the number of people who have birthdays on day number $i$ for
$1 \leq i \leq b.$
We approximate the joint distribution of $(X_1, \ldots, X_b)$ by assuming that each $X_i$ is a Poisson variable with expected value $n/b,$
and assuming that these variables are all independent.

The probability that no $k + 1$ people all have their birthdays on day $i$ is
$P(X_i \leq k) = F(k),$ where $F$ is the CDF of $\mathrm{Poisson}(n/b).$
The probability that this happens on all $b$ days of the year, that is,
there is no day on which more than $k$ people have a birthday,
is $(F(K))^b.$

The estimated probability that there is a day when $k+1$ or more people have a birthday is therefore $1 - (F(K))^b.$

Setting $b=365$ and trying this out on a few known cases
from https://oeis.org/A014088:

\begin{array}{ccccc}
k & n & n/b & F(k) & (F(k))^b \\
1 & 22 & 0.0602740 & 0.99825489 & 0.5286 \\
1 & 23 & 0.0630137 & 0.99809610 & 0.4988 \\
2 & 87 & 0.2383562 & 0.99811045 & 0.5014 \\
2 & 88 & 0.2410959 & 0.99804850 & 0.4902 \\
3 & 186 & 0.5095890 & 0.99812428 & 0.5039 \\
3 & 187 & 0.5123288 & 0.99808774 & 0.4973 \\
4 & 312 & 0.8547945 & 0.99812038 & 0.5032 \\
4 & 313 & 0.8575342 & 0.99809433 & 0.4985 \\
\end{array}

This predicts (correctly) that you need $23$ people to have at least a $50\%$ chance of at least one set of two people with the same birthday,
$88$ people to have at least a $50\%$ chance of at least one set of three people with the same birthday,
$187$ people to have at least a $50\%$ chance of at least one set of four people with the same birthday,
$313$ people to have at least a $50\%$ chance of at least one set of five people with the same birthday.

As another example, suppose you have $100$ people in the room. Then this approximation gives $(F(2))^{365} \approx 0.3600$,
and therefore the probability of three or more people all with the same birthday is approximately $0.6400.$
Wolfram Alpha
gives the probability as $0.6459$.
Contrast this with the accepted answer, which estimates the probability at $0.7029.$

On the other hand, for small numbers of people in the room, this method overestimates the chance of $k+1$ people with the same birthday.
For example, for the probability of three or more people with the same birthday out of $30$ people, this method gives $0.0313$, the accepted answer gives $0.0300,$ and Wolfram Alpha gives $0.0285.$

The individual $X_i$ are not actually Poisson and of course they are not actually independent.
That is what makes this an approximation rather than an exact method.