ROC of this LTI system given $x[n]$ and $y[n]$
Let's look at the pole locations and the corresponding regions of convergence (ROCs) of the $\mathcal{Z}$-transforms of the input and output signals:


$X(z)$: poles at $z=\frac12$ and $z=2$ with ROC $\frac12&lt;|z|&lt;2$ (two-sided).
$Y(z)$: poles at $z=\frac12$ and $z=\frac34$ with ROC $|z|&gt;\frac34$ (causal).


Now we know that the ROC of the multiplication of two $\mathcal{Z}$-transforms equals the intersection of the ROCs of the individual $\mathcal{Z}$-transforms, unless there is pole-zero cancellation, in which case the resulting ROC can be bigger than the intersection.

$H(z)$ has a zero at $z=2$ and a pole at $z=\frac34$. Since the zero cancels the pole of $X(z)$, the restriction $|z|&lt;2$ for the ROC of $Y(z)$ is removed. The two theoretically possible ROCs of $H(z)$ are $|z|&lt;\frac34$, corresponding to an anti-causal and unstable system, and $|z|&gt;\frac34$, corresponding to a causal and stable system. In the first case, the ROC of $Y(z)$ would need to be $\frac12&lt;|z|&lt;\frac34$, which is not the case, and which would correspond to a two-sided unstable signal. $^1$ Consequently, the ROC of $H(z)$ must be $|z|&gt;\frac34$, which means that the system described by $H(z)$ is causal and stable.


1. I use the term unstable signal for a signal that, if interpreted as the impulse response of an LTI system, would correspond to an unstable system (i.e., the ROC does not contain the unit circle).