Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ 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?

Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ 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 $$\mathbb{E}(Z^p)=\frac{(\sigma_y/\sigma_x)^2}{{M-1}\choose{p}}.$$

This answer looks simple enough, I wonder which distribution $P(Z)$ has reciprocals of binomial coefficients as its moments?

Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ 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 $$\mathbb{E}(Z^p)=\frac{p!\Gamma(M-p)}{\Gamma(M)}(\sigma_y/\sigma_x)^2.$$ You will want to check this answer, an error is easily made.

Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ 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?