You are counting some cases more than once. Example:

For three persons, $A,B,C$ and for January 1st you have:

$A,B$ born on 1.1: $\frac{1}{365^2}$

$B,C$ born on 1.1: $\frac{1}{365^2}$

$A,C$ born on 1.1: $\frac{1}{365^2}$

Which are all true, but you counted the case that $A,B,C$ were all born on 1.1 three times.

This is called the inclusion-exclusion principle, and can indeed be employed to solve the birthday problem.

However, there is a much simpler solution if you consider the probability that all birthdays are on distinct days and take its complement.

You are counting some cases more than once. Example  :

For three persons, $A,B,C$ and for January 1st you have:

$A,B$ born on 1.1: $\frac{1}{365^2}$

$B,C$ born on 1.1: $\frac{1}{365^2}$

$A,C$ born on 1.1: $\frac{1}{365^2}$

Which are all true, but you counted the case that $A,B,C$ were all born on 1.1 three times.

This is called the inclusion-exclusion principle, and can indeed be employed to solve the birthday problem.

However, there is a much simpler solution if you consider the probability that all birthdays are on distinct days and take its complement.