First here, by the Cauchy--Schwarz inequality,
$$E|z_1z^*_2|=E|z_1|\,|z^*_2|\le\sqrt{E|z_1|^2E|z^*_2|^2}=E|z_1|^2<\infty,$$
by Addition in response to the modification of the OP's original question.
So, $Ez_1z^*_2$ exists and is finite.
Therefore and because the joint distribution of the pair $(-z_1,z^*_2)$ is the same as that of $(z_1,z^*_2)$, we conclude that
$$Ez_1z^*_2=E(-z_1)z^*_2=0.$$

**Detail sdded in response to the OP's comment:** The $x_i$'s are iid and, for each $i$, the distribution of $-x_i$ is the same as that of $x_i$. So, the joint distribution of $(-x_1,x_2,\dots,x_M)$ is the same as that of $(x_1,x_2,\dots,x_M)$. Also, $|-x_1|=|x_1|$. So, the random pair
\begin{multline*}
(z_1,z^*_2)=g(x_1,x_2,\dots,x_M):= \\
\Big(\frac{x_{1}}{|x_{1}|^2 + |x_{2}|^2 + \dots + |x_{M}|^2},
\frac{x_{2}}{|x_{1}|^2 + |x_{2}|^2 + \dots + |x_{M}|^2}\Big)
\end{multline*}
equals
\begin{multline*}
(-z_1,z^*_2)=
\Big(\frac{-x_{1}}{|-x_{1}|^2 + |x_{2}|^2 + \dots + |x_{M}|^2},
\frac{x_{2}}{|x_{1}|^2 + |x_{2}|^2 + \dots + |x_{M}|^2}\Big) \\
=g(-x_1,x_2,\dots,x_M)
\end{multline*}
in distribution,
as was stated above.