Well, the probability of having at least $1$ person with the birthday
of $q$ is $n/365$.
Unfortunately, this isn't quite true. If you could guarantee that the $n$ people didn't share a birthday, then it would be true, but if you could guarantee that, there wouldn't be much of a birthday question! The actual probability that at least one person out of $n$ has $q$ as his or her birthday is
$$
1-\left(\frac{364}{365}\right)^n
$$
where we assume (as is typical for initial analyses of this problem) that birthdays are uniformly distributed over a $365$-day year. If $n$ is small, this is approximately equal to $n/365$ (and it is exactly true for $n = 1$), but this is increasingly untrue as $n$ becomes larger.
Note that if $n > 365$ then $n/365 > 1$, but we know it is perfectly possible for each of (let's say) $400$ people to happen not to have May $21$ for their birthday. The above expression gives that probability (usual assumptions applying).
To get the probability of $2$ people, you would multiply the
probability of having one person by $(n−1)/365$.
This also isn't true. In order to determine the probability that at least two people have $q$ as their birthday, it's simplest to subtract from the above probability the chance that exactly one person has $q$ as their birthday. This yields
$$
1-\left(\frac{364}{365}\right)^n
-n\left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{n-1}
$$
As before, what you say is approximately true for small $n$ (and exactly true for $n = 2$), but still is increasingly untrue for larger $n$, for nearly the same reason.
And if you wanted the probability of at least $k$ of the $n$ people sharing $q$ as their birthday, we could continue subtracting all the way up to $k-1$ people sharing $q$ as their birthday:
$$
1-\sum_{i=0}^{k-1} \binom{n}{i} \left(\frac{1}{365}\right)^i \left(\frac{364}{365}\right)^{n-i}
$$
So we can say, given $n$ people, the probability that $2$ of them will
both have the birthday $q$ (and therefore the same birthday) is $(n^2-n)/365^2$.
Even if the above were true, the analysis would have to be expanded to cover all possible $q$, rather than only a single $q$. If you evaluate my second expression above at $n = 2$, the result is $1/365^2$, the probability that two people share $q$ as their birthday. If you multiply that by $365$, the number of possible choices for $q$, you get $1/365$, which is the probability that two people share any day as their birthday. However, that expression cannot be extended to expand beyond $n = 2$ exactly, because it neglects the possibility that more than two people share the same birthday.
The standard analysis of the birthday problem is the most straightforward, in which the probability of at least two of $n$ people sharing a birthday is
$$
1-\left(\frac{365}{365}\right)
\left(\frac{364}{365}\right)
\left(\frac{363}{365}\right) \cdots
\left(\frac{366-n}{365}\right)
$$
As far as I can tell, your methods do not work and cannot be salvaged.
This is a hard problem that cannot be solved by conventional means. To find the probability that at least three people have a common birthday, let us instead compute the probability that no three people share a birthday. This is equal to
$$
\frac{n^{365}}{365^n}.\tag 1
$$
Here, $[x^n]f(x)$ is the coefficient of $x^n$ in the polynomial $f(x)$. You want one minus the above. Another way to write this is
$$
\sum_{a_1,a_2,\dots,a_{365}}\frac1{365^n}\frac{n!}{a_1!a_2!\cdots a_{365}!}.\tag 2
$$
where the sum ranges over all vectors $(a_1,a_2,\dots,a_{365})$ of integers between $0$ and $2$ whose sum is $n$. This works because each vector specifies a valid distribution of birthdays where no birthday repeats three times or more, and the multinomial coefficient gives the number of ordered selections of birthdays which have that distribution, and each is then multiplied by the probability $(1/365)^n$ of that ordered selection occurring. You can check that when you expand out $(1)$ and collect the coefficient of $x^{n}$, you get exactly $(2)$.
If you are okay with an approximate result, the expected number of triplets of people who share a common birthday is $\lambda =\binom{n}{3}\cdot \frac1{365}^2$, and the number of triplets with a common birthday is approximately Poisson with parameter $\lambda$. Therefore, the probability some triplet share a birthday is approximately
$$
1-e^{-\binom{n}3/365^2}.
$$
Best Answer
I will leave you to judge whether this is the same answer as yours:
With $d=365$ days, the probability a particular pair share a birthday is $\frac1{d}$
With $n$ people, there are ${n \choose 2}= \frac{n(n-1)}{2}$ pairs of people so the expected number of pairs sharing a birthday is $\frac{n^2-n}{2d}$
This is greater than $1$ when $n > \frac{1 +\sqrt{1+8d}}{2}$ or $n < \frac{1 -\sqrt{1+8d}}{2}$, though for this question the answer should be positive so $k \gt 27.523\ldots$ and an integer
Note that you might get a slightly different answer if your question was looking for the expected number of days to be shared to be at least $1$:
The probability a particular day is shared is $1-\frac{(d-1)^n}{d^n} - n\frac{(d-1)^{n-1}}{d^n} $
The expected number of shared days is $d-(n+d-1)\left(1-\frac{1}{d}\right)^{n-1}$
For the same $k$ and $d$ this expected number is slightly smaller than the earlier expression, as three people and so three pairs might share a single day.
It is also harder to treat analytically, but here it will be greater than $1$ when $k \gt 28.175\ldots$, not surprisingly marginally more than the original version of the question