Your work is fine so far! However, the question states that we only consider a single observation. Therefore, you just have to consider the quotient of the likelihoods for one observation $x$: The critical region $C$ is given by
$$\frac{L(\theta_0 \ | \ x)}{L(\theta_1 \ | \ x)} \geq k.$$
An easy calculation gives
$$\frac{L(\theta_0 \ | \ x)}{L(\theta_1 \ | \ x)} = \frac{\theta_0 (1-\theta_0)^{x-1}}{\theta_1 (1-\theta_1)^{x-1}} = \left(\frac{\theta_0}{\theta_1}\right)\left(\frac{1-\theta_0}{1-\theta_1}\right)^{x-1} \geq k,$$
which implies that (please recheck this)
$$x \geq 1 + \frac{\ln\left(\frac{k \theta_1}{\theta_0}\right)}{\ln\left(\frac{1-\theta_0}{1-\theta_1}\right)} := k^{*}.$$
Hence, by the Neyman-Pearson lemma, the rejection region for the most powerful hypothesis test $H_0: \theta = \theta_0$ and $H_A: \theta =\theta_1$ where $\theta_1>\theta_0$ is given by $x \geq k^{*}$. Note that since the geometric distribution is discrete, this critical region $C = \{k^{*},k^{*}+1,\ldots,\}$. We still need to compute $k^{*}$. This can be done by looking at the type $I$-error, since $\mathbb{P}(H_0 \ \mbox{is false} \ | \ H_0) = \alpha$. Now $H_0$ is false if $x \geq k^{*}$ and hence the type $I$-error satisfies
\begin{align}
\mathbb{P}(X \geq k^{*} \ | \theta = \theta_0) = \sum_{k = k^{*}}^{\infty} (1-\theta_0)^{k-1} \theta_0 = \alpha,
\end{align}
from which you can extract $k^{*}$. I think you can also generalize this to multiple observations $x_1,\ldots,x_n$. In that case you will have to determine the distribution of $\overline{x}$ which can be a little bit more messy.
