Calculating the output of a pole eigen signal in a difference equation
If the input is $x[n]=\left(-\frac12\right)^n$ for all $n$, then the output is infinite, because the convolution sum doesn't converge. Note that also the input grows without bounds for negative $n$.
If, on the other hand, the input starts at $n=0$, i.e., $x[n]=\left(-\frac12\right)^nu[n]$ then you can use the $\mathcal{Z}$-transform to compute the output:
$$Y(z)=X(z)H(z)\tag{1}$$
With $$X(z)=\frac{1}{1+\frac12 z^{-1}}\tag{2}$$
you obtain
$$\begin{align}Y(z)&amp;=\frac{1+3z^{-1}}{\left(1+\frac12 z^{-1}\right)^2\left(1-\frac12 z^{-1}\right)}\\&amp;=\frac{a}{\left(1+\frac12 z^{-1}\right)^2}+\frac{b
}{1+\frac12 z^{-1}}+\frac{c}{1-\frac12 z^{-1}}\tag{3}\end{align}$$
with some real-valued constants $a$, $b$, and $c$.
Inverse $\mathcal{Z}$-transform of $(3)$ gives
$$y[n]=-2a\;n\left(-\frac12\right)^n+b\left(-\frac12\right)^n+c\left(\frac12\right)^n\tag{4}$$

EDIT: First, to clarify, I use a basic example that shows that a double pole in the $\mathcal{Z}$-transform is no problem. Assume a system with impulse response $h[n]=(1/2)^nu[n]$ and an input signal $x[n]=(1/2)^nu[n]$. Computing the output in the time domain is straightforward:
$$\begin{align}y[n]&amp;=u[n]\sum_{k=0}^n\left(\frac12\right)^k\left(\frac12\right)^{n-k}\\&amp;=u[n]\left(\frac12\right)^n\sum_{k=0}^n1\\&amp;=(n+1)\left(\frac12\right)^nu[n]\tag{5}\end{align}$$
The $\mathcal{Z}$-transform of the output sequence is simply given by the multiplication of the $\mathcal{Z}$-transforms $H(z)$ and $X(z)$:
$$Y(z)=\frac{1}{\left(1-\frac12z^{-1}\right)^2}\tag{6}$$
The inverse $\mathcal{Z}$-transform of $(6)$ equals the result $(5)$, which is very easy to confirm.

The probable cause of the OP's confusion could be the fact that for an input signal $x[n]=z_0^n$, $-\infty&lt;n&lt;\infty$, the output is given by
$$y[n]=H(z_0)z_0^n$$
if and only if $z_0$ is inside the ROC of $H(z)$, because otherwise the convolution sum doesn't converge and the expression $H(z_0)$ is meaningless. However, in the computation of the outputs $y[n]$ in the examples above we never evaluate $H(z)$ for any $z$ outside the ROC, so the situation is entirely different.