Let $\{Z_t\}$ be a sequence of random variables that are weakly stationary, i.e., $\mathbb{E}[Z_t]=0$ and $Var[Z_t]=\sigma^2$ and $Z_t$'s are uncorrelated, i.e., $cov(Z_s,Z_t)=0$ for all $t\neq s$. Can we create such a sequence of $Z_t$ that are not independent? Also, can we have the other way around, that is, $Z_t$'s are independent but not identically distributed?
My try: For the first question, I am thinking of $Z_t = (-1)^tX_t$ where $X_t \sim N(0,1)$ but do not know how to show they are dependent.
If one chooses to define discrete-time white noise $\{Z_n\}$ as a sequence of zero-mean random variables that are uncorrelated but not necessarily independent, then there is nothing left to prove; the definition says it all. The $Z_n$'s are uncorrelated but need not be independent.
A far superior deinition of discrete-time white noise is as a sequence of zero-mean random variables that are independent (preferably identically distributed as well). The key point here is that with uncorrelated but dependent random variables, one cannot make a linear prediction (other than $0$) of the value of $Z_{n+i}$ given $Z_n$ (say) but it might be possible to make a nonlinear prediction. In fact, the minimum mean-square error predictor of $Z_{n+i}$ when we know $Z_n$ is $E[Z_{n+i}\mid Z_n]$ which is, in general, a nonlinear function of $Z_n$. It sticks in my craw that one should be able to predict the value of white noise in any way (not just linear prediction), and the requirement of independence for white noise sequences eliminates this possibility: if $Z_{n+i}$ and $Z_n$ are independent, then $E[Z_{n+i}\mid Z_n] = E[Z_{n+i}] = 0$.
Now, if one wants a sequence of zero-mean random variables that are uncorrelated but not necessarily independent, consider the following.
Let $\{X_{2n}\colon n\in \mathbb Z\}$ be a sequence of independent identically distributed zero-mean standard normal random variables, and let $\{B_n \colon n\in \mathbb Z\}$ be a sequence of independent discrete random variables taking on values $+1$ and $-1$ with equal probability $\frac 12$ and independent of the $X_{2n}$'s. Now set $X_{2n+1} = X_{2n}B_n$ for each integer $n\in \mathbb Z$. Verify that $X_{2n+1}$ is also a standard normal random variable and that $X_{2n}$ and $X_{2n+1}$ are uncorrelated normal random variables (hint: $E[X_{2n}X_{2n+1}] = E[X_{2n}^2]E[B_n] =0$). But note that $X_{2n}$ and $X_{2n+1}$ are not independent random variables; indeed, conditioned $X_{2n}=\alpha$ we know that the conditional distribution of $X_{2n+1}$ is that of a discrete random variable taking on values $\pm\alpha$ with equals probability instead of serenely continuing to be standard normal as independence demands. Thus, $\{X_{m}\colon m\in \mathbb Z\}$ is a sequence of uncorrelated standard normal random variables (discrete-time white noise process according to the OP's definition) in which pairs $(X_{2n},X_{2n+1})$ are not independent (though they are uncorrelated). Call such pairs doublets. More generally, $(X_m, X_k)$ are either part of the same doublet (and hence uncorrelated but not independent) or belong to different doublets and thus are uncorrelated as well as independent.
Turning to the OP's other questions to whether the $Z_n$'s can be independent but not identically distributed, it is trivial to create such a "white noise process" according to the OP's definition. All we need is to have a sequence of independent zero-mean random variables with varying distribution functions, but all having the same finite variance $\sigma^2$; for example. $Z_n \sim N(0,\sigma^2)$ if $n$ is odd and $Z_n \sim U\left(-\sqrt{3}\sigma. +\sqrt{3}\sigma\right)$ if $n$ is even.It is not clear to me that this type of process has any particularly important role to play in any situation, but ymmv just as much as the OP's seems to be doing.