Intuition about independent signals
Think it this way; assuming $x[-1] = 0$, then recusively compute the output $x[k]$ for $k \ge 0$ such as

$$\begin{align}
x[0] &amp;= v[0] \\
x[1] &amp;= a x[0] + v[1] \\
x[2] &amp;= a x[1] + v[2] = a^2 x[0] + a v[1] + v[2] = a^2 v[0] + a v[1] + v[2] \\
x[2] &amp;= \left( a^2 v[0] + a v[1]  \right) + v[2] \\
... &amp;= ... \\
x[k] &amp;= \left( a^k v[0] + a^{k-1}v[1] + ... + a v[k-1] \right) + v[k] \\
x[k] &amp;= a ~ x[k-1] + v[k] \\ 
\end{align}
$$ 

As you can see at the bottom two lines, the output at time $k$ uses all input samples up to time $k-1$ and the new input sample $v[k]$. Since $v[k]$ is declared as an independent random process, then all of its random variables $v[0],v[1],v[2],...,v[k]$ are independent from each other. Then the current input $v[k]$ will be independent from the past output $x[k-1]$, as that output does not contain any contributions from the input sample $v[k]$;

So what's the illusion in this equation 
\begin{equation}
x[k-1]=\frac{x[k]-v[k]}{a}
\end{equation}

which seems to suggest a dependency between $x[k-1]$ and $v[k]$ ?

If you consider the causal solution of the difference equation as outlined recursively above, then it's seen that the current output sample $x[k]$ contains a contribution from current input sample $v[k]$ and plus all past input contributions from $v[0],v[1],...,v[k-1]$ but then the difference $x[k]-v[k]$ will cancel out the presence of the current input $v[k]$ at the right hand side of the equation. Hence the left hand side $x[k-1]$ will be actually depending on the past inputs alone and not on the current input (which apprently explicitly shows up there but is actually implicitly cancelled by the difference of $x[k]$ and $v[k]$).