Variance Due to white noise input
You have a typo in your definition of $R_y(k)$, and an error in the time indices when developing the equation. The proper definition of autocorrelation for complex signals (in the case of wide-sense stationary processes) is
$$R_y(k)=\mathbb{E}\left[y(n)\overline{y(n+k)}\right],$$
and setting $k=0$, we obtain
$$\begin{align}
R_y(0)&amp;=\mathbb{E}\left[y(n)\overline{y(n+0)}\right]=\mathbb{E}\left[\left|y(n)\right|^2\right]\\
&amp;=\mathbb{E}\left[\left|\sum_{m=-\infty}^{\infty}h(m)x(n-m)\right|^2\right]\\
&amp;=\mathbb{E}\left[\left(\sum_{m=-\infty}^{\infty}h(m)x(n-m)\right)\overline{\left(\sum_{i=-\infty}^{\infty}h(i)x(n-i)\right)}\right]\\
&amp;=\mathbb{E}\left[\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}x(n-m)\overline{x(n-i)}\right]\\
&amp;=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,\mathbb{E}\left[x(n-m)\overline{x(n-i)}\right]\\
&amp;=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\
\end{align}$$
where I first use the fact that the product of sums is the sum of products (and I use a different summation index to better show the cross-products), and second, that the impulse response of the system is deterministic and therefore constant for the purposes of taking expectation.
Now, we can group those terms where $m=i$ and those where $m\neq i$, obtaining
$$\begin{align}
R_y(0)&amp;=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\
&amp;=\sum_{m=-\infty}^{\infty}h(m)\overline{h(m)}\,R_x(0)+\sum_{m=-\infty}^{\infty}\sum_{i=-\infty\\ i\neq m}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\
&amp;=R_x(0)\sum_{m=-\infty}^{\infty}|h(m)|^2+\sum_{m=-\infty}^{\infty}\sum_{i=-\infty\\ i\neq m}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\
\end{align}$$
Finally, we know that the input noise is white, and therefore $R_x(k)=0$ for $k\neq0$. We also know that $R_x(0)=\sigma_x^2$. Thus, the second term vanishes, and we obtain
$$\begin{align}
R_y(0)
&amp;=R_x(0)\sum_{m=-\infty}^{\infty}|h(m)|^2=\sigma_x^2\sum_{m=-\infty}^{\infty}|h(m)|^2
\end{align}.$$