Why use a Z test rather than a T test for confidence interval of a population proportion?
Let's forget about population proportions for a second. Let's say we are placing a confidence interval on the population mean of some random variable X. My understanding is that if the variance of X is known, then we can do a Z test. Otherwise (the common case), we must estimate the variance from the sample, and so we should do a T test. My understanding is that this is true EVEN if X is normally distributed. That is, if we estimate variance from the sample, the sampling distribution (not sure I have this part quite right) is a T distribution of n-1 degrees of freedom, even if X is normally distributed.
Why does the same logic not apply to estimating the population proportion? In online textbooks  & videos , a Z test is being done instead. My understanding is that if the sample size is large, the binomial distribution can be approximated with a normal distribution due to the central limit theorem, but even if that is so, aren't we estimating variance from the sample, implying the need for a T test, not a Z test?