If you just want to check whether the $\mathcal{Z}$-transform of a sequence exists, it is not necessary to actually compute its $\mathcal{Z}$-transform. Fat32's answer showed you how to do that for your example.
The $\mathcal{Z}$-transform of a sequence $x[n]$ exists if for a set of points in the complex plane
$$\left|\sum_{n=-\infty}^{\infty}x[n]z^{-n}\right|<\infty\tag{1}$$
is satisfied. Note that in the formula in your question you forgot the magnitude, which is essential.
For your example it is straightforward to show that $(1)$ is satisfied for $|z|>1$:
$$\begin{align}\left|\sum_{n=-\infty}^{\infty}x[n]z^{-n}\right|&=\left|\sum_{n=0}^{\infty}\sin(\omega_0n)z^{-n}\right|\\&\le \sum_{n=0}^{\infty}\big|\sin(\omega_0n)\big|\;|z|^{-n}\\&\le\sum_{n=0}^{\infty}|z|^{-n}=\frac{1}{1-|z|^{-1}},\qquad |z|>1\tag{2}\end{align}$$
where I've used $|\sin(\omega_0n)|\le 1$.
Eq. $(2)$ shows that the $\mathcal{Z}$-transform of the given sequence exists for $|z|>1$. The region $|z|>1$ in the complex plane is called region of convergence (ROC).