I've seen a couple similar questions on stack exchange but did not find their answers to be enlightening. Perhaps the issue traces back to a confusion with the Bernoulli and Binomial distributions. (Sidebar, can the Binomial population variance be determined via the formula np(1-p) or is this just the sample variance?)
It's my understanding that the t-test is preferred when population variance is unknown, (thus the sample variance must be used instead) when dealing with sample means.
Additionally, the t-test is appropriate when samples are drawn from normal (potentially approximately normal?) distributions. In the case of proportions, the distribution is defined on the range [0,1] and so the assumption of normality does not hold.
However, due to the central limit theorem, the means sampled from a binomial distribution will be approximately normal when n is large enough, so the z-test can be used. But, the t-distribution converges on the normal distribution when n and degrees of freedom are large enough. And in the case of proportions, n is usually very high (certainly above 30 for commercial AB tests, anyway.)
So, it would seem that the justification is more practical than proof/theory based; the t-test is just a pain to use and since n is so high, why not just use a z-test? However, in one of the questions, user whuber, addressed this specifically in a comment, however, I didn't understand his argument.
