Using ROC to find stability of system in specific example
because even though it has more poles than zeros


WRONG. The Z-transform transfer function will always have equal number of poles and zeros. Your poles are at $z = -5/4$ and $z = 1/4$. Zeros are at $z = -1/2$ and $z = \infty$. Since ROC will always be concentric circle region without including poles, there are 3 possible ROC for the given transfer function.
$$
ROC_1 = 5/4 \lt |z| \lt \infty
$$
$$
ROC_2 = 1/4 \lt |z| \lt 5/4
$$
$$
ROC_3 = 0 \le |z| \lt 1/4
$$
$ROC_1$ represents an unstable but causal system (since it can include $z \rightarrow \infty$ so terms like $z^{-k} \,\,with\,\, k \ge 0$ will only be present (only negative or zero powers of $z$). And because only negative powers of $z$ are present, system is non-zero for $n \ge 0$. So it is causal. Example - $H(z) = a_0 + a_1z^{-1}+a_2z^{-2}+a_3z^{-3}+a_4z^{-4}+...$

$ROC_2$ represents stable but non-causal system because it includes unit circle $|z| = 1$ but does not include $z \rightarrow \infty$. Terms with $z$ raised to positive power can be there along with negative power of $z$. So you can rule out $z=0$ and $z \rightarrow \infty$.Example - $H(z) = a_0 + a_1z^{-1}+a_2z^{-2}+a_3z^{-3}+a_4z^{-4}+a_{-1}z^{1}+a_{-2}z^{2}+a_{-3}z^{3}$

$ROC_3$ represents an unstable and non-causal system because it does not include unit circle and terms with only positive power of $z$ are present (because ROC includes $z=0$ and does not include $z \rightarrow \infty$. Example - $H(z) = a_0 + a_{-1}z^{1}+a_{-2}z^{2}+a_{-3}z^{3}+...$

In your case, the ROC might be $ROC_3$.