z-test assumptions

A z-test can be performed instead of a t-test when the sample size is large since the t-distribution with many degrees of freedom is similar to the standard normal distribution. In this case I understand that all assumptions for a t-test must be met. This including data from a normally distributed variable.

But a z-test can also be performed when the population standard deviation is known. According to the central limit theorem the sampling distribution will be normally distributed if the sample size is large. Must also the variable be normally distributed in this case?

In other words: a normally distributed sampling distribution in not sufficient for the t-score to be t-distributed but is it sufficient for the z-score to be standard normal in the case where the standard deviation i known? If I have data from a skewed distribution with known standard deviation, can I use a z-test if the sample size is large?

I know that it's an odd situation but the answer will hopefully help me understand why a normally distributed sampling distribution not is sufficient for the t-test.