Confidence intervals are typically formed by finding a pivotal quantity that involves the parameter of interest, writing a probability statement for that pivotal quantity falling within an interval, and then "inverting" that probability statement to get it in terms of an interval for the parameter. As a first step, you should be trying to come up with a quantity $f(\hat{\theta}, \theta)$ that has a fixed distribution that does not depend on the parameter:
$$f(\hat{\theta}, \theta) \overset{\text{Approx}}{\sim} \text{Fixed Dist}.$$
Have a think about how the central limit theorem might be applied here to get an approximate distribution for an appropriately standardised version of $\bar{x}$. This ought to lead you to a pivotal quantity, which you can then use to write a probability statement for your confidence interval. (You can find some examples of the derivation of confidence intervals in this related answer).
Updated based on your proposed solution: Using the CLT, for large $n$ we have:
$$\hat{\theta} = \frac{2}{3} \bar{X}_n \overset{\text{Approx}}{\sim} \text{N} \bigg( \theta, \frac{12 \theta^2}{n} \bigg).$$
You can use this to form a quasi-pivotal quantity with a standard normal distribution and then form a confidence interval using that quantity. That should give you a confidence interval of the stipulated form, but it's not the best way to do this. Instead, I would recommend building the confidence interval with the MOM estimator using the quasi-pivotal quantity:
$$n \cdot \frac{(\hat{\theta} - \theta)^2}{12 \theta} \overset{\text{Approx}}{\sim} \text{ChiSq}(\text{df} = 1).$$
You can now derive the confidence interval in a manner analogous to the Wilson score interval. To facilitate this analysis, let $\chi_{1,\alpha}^2$ denote the critical point of the chi-squared-one distribution with upper tail area $\alpha$. For any confidence level $1-\alpha$ we can write the probability interval:
$$\begin{align}
1-\alpha
&\approx \mathbb{P} \bigg( n \cdot \frac{(\hat{\theta} - \theta)^2}{12 \theta} \leqslant \chi_{1, \alpha}^2 \bigg) \\[6pt]
&= \mathbb{P} \bigg( n (\hat{\theta} - \theta)^2 \leqslant 12 \chi_{1, \alpha}^2 \theta \bigg) \\[6pt]
&= \mathbb{P} \bigg( n \Big[ \theta^2 - 2 \Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big) \theta + \hat{\theta}^2 \Big] \leqslant 0 \bigg) \\[6pt]
&= \mathbb{P} \bigg( \Big( \theta - \hat{\theta} - 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 \leqslant \Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n} \bigg) \\[6pt]
&= \mathbb{P} \bigg( \Bigg| \theta - \hat{\theta} - 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Bigg| \leqslant \sqrt{\Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n}} \bigg) \\[6pt]
&= \mathbb{P} \bigg( \theta \in \bigg[ \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \pm \sqrt{\Big( \hat{\theta} + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}^2}{n}} \bigg] \bigg). \\[6pt]
\end{align}$$
Upon substitution of the actual data into the estimator, you get the confidence interval:
$$\text{CI}_\theta(1-\alpha) \equiv \bigg[ \hat{\theta}_n + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \pm \sqrt{\Big( \hat{\theta}_n + 6 \cdot \frac{\chi_{1, \alpha}^2}{n} \Big)^2 - \frac{\hat{\theta}_n^2}{n}} \bigg].$$
This is probably about as good as you're going to get based on the MOM estimator. However, it's worth noting that this is a pretty crappy confidence interval estimator in this case. In cases where the parameter of interest determines the bounds of the support, it is almost always better to use an estimator based on the extreme order statistics rather than the mean.
