BIBO stability and input at system's pole
The signal $x[n] = (\frac{1}{3})^n$ is not bounded. As $n \to -\infty$, $x[n] \to \infty$; therefore, the system's response to your candidate signal $x[n]$ has no bearing on its BIBO stability.

A bounded, causal version of the signal (and likely what you were thinking) is $x_b[n] = (\frac{1}{3})^n u[n]$, where $u[n]$ is the unit step function. The response of the system to this signal is bounded as you would expect.