Using the representation in this <A HREF="https://mathoverflow.net/a/323376/11260">answer</A>, 
$$Z=  \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$
and integrating over
the independent chi-squared variables $\xi_2$, $\xi_{2M}$, and $\xi_{2M-2}$ I find the expectation value
$$\Rightarrow\mathbb{E}(Z)=\frac{1}{M-1}(\sigma_y/\sigma_x)^2.$$
More generally, the $p$-th moment is finite for $M>p$, given by
$$\mathbb{E}(Z^p)=\frac{(\sigma_y/\sigma_x)^2}{{M-1}\choose{p}}.$$
You will want to check this answer, an error is easily made.

This answer looks simple enough, I wonder which distribution has reciprocals of binomial coefficients as the moments?