To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:

Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace inside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x = \sigma_y^2 \, \Vert x \Vert^2$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$

To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:

Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace outside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x = \sigma_y^2 \, \Vert x \Vert^2$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$

(This might also be helpful in other situations, since normality is not used until the very end.)

To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:

Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace inside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$

To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:

Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace inside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x = \sigma_y^2 \, \Vert x \Vert^2$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$