This is a first-order-real-valued autoregressive (AR) process $y(n)$ that satisfies the real-valued difference equation $y(n)+a_1y(n-1)=x(n)$ where $a_1$ is a constant and x(n) is a white noise process with variance $\sigma_x^2$
Although x[n] is a white noise, I have to assume that it has a non-zero mean, I am assuming that at least it is WSS, hence $E\{x[n]\}=\mu_x [n]=\mu_x\not=0$.
Now the filter equation would be $H(Z)=\frac{1}{1+a_1z^{-1}}\leftrightarrow h[n]=(-a_1)^nu[n]$, then $\mu_y[n]=\mu_y=\mu_x\sum\limits_{n=-\infty}^\infty (-a_1)^nu[n]=\mu_x\sum\limits_{n=0}^\infty (-a_1)^n=\frac{\mu_x}{1+a_1} $ (Geometric Serie)
I have also found that its $R_{xx}(n_1n_2)=R_{xx}(l)=(\sigma_x^2+\mu_x^2)\delta(n)$ for $l=n_1-n_2=0$ and $S_{xx}(Z)=\sigma_x^2+\mu_x^2$. Then, I am getting that $S_{yy}(z)=H(z)H(z^{-1})S_{vv}(z)=\frac{\sigma_x^2+\mu_x^2}{(1+a_1z)(1+a_1z^{-1})}\leftrightarrow R_{yy}(l)=\frac{(\sigma_x^2+\mu_x^2)a_1^{|n|}}{1-a_1^2}$
I am getting that it depends on $l$, I am not sure of assuming that x[n] has a non-constant mean or that it is not WSS, but with this procedure,the output y[n] is WSS, I am not sure how to check that is SSS
There is an important assumption missing in your development: $x[n]$ is assumed to be causal, i.e., $x[n] = 0$ for $n \lt 0$. This is directly related with the customary initial conditions we define to solve the difference equation of the AR process.
We then first convolve $h[n]$ with $x[n]$ to get an expression of $y[n]$ in terms of $x[n]$:
(1) $y[n] = h[n] * x[n] = \sum_{k=0}^n (-a_1)^k x[n-k]$
And after that we take the expected value operation at both sides of (1):
\begin{align} E\{y[n]\} &= \sum_{k=0}^n (-a_1)^k E\{x[n-k]\}\\ &= \mu_x\sum_{k=0}^n (-a_1)^k\\ &= \left \{ \begin{array}{ll} \mu_x\frac{1-(-a_1)^{n+1}}{1+a} & a_1 \neq -1\\ (n+1)\mu_x & a_1 = -1 \end{array} \right . \end{align}
And since $\mu_x \neq 0$, $E\{y[n]\}$ depends on $n$.