Let $X_1,\ldots,X_n$ be a a sample from a Uniform Distribution $(0,\theta)$ where $\theta > 0$ is an unknown parameter.
I have found the estimator based on the sample mean
$$\hat{\theta}=2\bar{X}$$
As well as the estimator based on the nth order statistic
$$\hat{\theta}=\frac{n+1}{n}X_{(n)}$$
Which would be a better estimator in the MSE sense ?
Mean square error (MSE) of an estimator $\hat\theta$ for estimating $\theta$ is defined as $$\operatorname{MSE}_{\theta}(\hat\theta)=\operatorname E_{\theta}(\hat\theta-\theta)^2=\operatorname{Var}_{\theta}(\hat\theta)+(\text{bias}(\hat\theta))^2$$
Both your estimators are unbiased for $\theta$, so MSE here is just variance.
Now, $$\operatorname{Var}_{\theta}(2\overline X)=4\operatorname{Var}_{\theta}(\overline X)=4\times\frac{\theta^2}{12n}=\frac{\theta^2}{3n}$$
And keeping in mind that $X_{(n)}/\theta\sim \mathsf{Beta}(n,1)$ we have,
\begin{align} \operatorname{Var}_{\theta}\left[\frac{n+1}{n}X_{(n)}\right]&=\left(\frac{n+1}{n}\right)^2\operatorname{Var}_{\theta}(X_{(n)}) \\&=\left(\frac{n+1}{n}\right)^2\times\frac{n\theta^2}{(n+1)^2(n+2)} \\&=\frac{\theta^2}{n(n+2)} \end{align}
It is clear that
$$\frac{\theta^2}{3n}-\frac{\theta^2}{n(n+2)}=\frac{\theta^2}{n}\left[\frac{1}{3}-\frac{1}{n+2}\right]> 0\quad,\forall \,n> 1$$
This proves $T=\frac{n+1}{n}X_{(n)}$ has the smaller MSE, which is not surprising since it is known that $T$ is the uniformly minimum variance unbiased estimator of $\theta$.
As pointed out by @Henry, it is worth mentioning that $X_{(n)}$ is a sufficient statistic for $\theta$ (and hence responsible for data condensation without losing information about the parameter) whereas $\overline X$ is not. So even without any calculation we expect the estimator based on $X_{(n)}$ to have a smaller mean square error.