Not sure if I follow you, but as I understand your question you observed $k$ successes in $n$ trials and used this data in a beta-binomial model. If your prior was $\mathsf{Beta}(\alpha, \beta)$, then the posterior is $\mathsf{Beta}(\alpha + k, \beta + n-k)$. Now, let's go one step back, your model is
$$\begin{align} \mu &\sim \mathsf{Beta}(\alpha, \beta) \\ X_n &\sim \mathsf{Bin}(\mu, n) \end{align}$$
where $X_n = \sum_{i=1}^n Y_i$ for $Y_i \underset{i.i.d.}{\sim} \mathsf{Bern}(\mu)$. In plain English, you assume that $\mu$ is the probability of "success" for $n$ independent Bernoulli trials. The sum of the ones and zeros from the Bernoulli trials $X_n$ makes a binomial distribution. If now you want to make a guess on the number of successes in $m \ne n$ independent Bernoulli trials, you can just plug in $\mu$ into the binomial distribution $\mathsf{Bin}(\mu, m)$. The trials are identical and independent, so it is like tossing a coin $m$ times instead of $n$, each individual toss behaves the same regardless of how many times you toss.
You can also look at it through the lens of linearity of expectation. The expected value of a single Bernoulli trial is $E[Y_i] = \mu$, for $n$ trials it is $E[nY_i] = n E[Y_i] = n\mu$ and for $m$ trials $m\mu$. Those correspond to the means of the binomial distributions.
So yes, you just plug in the estimated $\mu$.