Some simplification is possible. Note that if $X_m$ is known, then consider the transformation $Y = X/X_m$, hence $Y$ is Pareto with shape $\alpha$ and location $1$ with density $f_Y(y) = \alpha y^{-(\alpha+1)} \mathbb 1 (y \ge 1)$. Then the transformed sample is simply $$\boldsymbol Y = (y_1, y_2, \ldots, y_n) = \boldsymbol X/X_m = (x_1/X_m, x_2/X_m, \ldots, x_n/X_m).$$ So for the purposes of estimation and efficiency, we can work with $\boldsymbol Y$, or equivalently, assume $X_m = 1$, because this transformation is one-to-one.
That said, the efficiency of an estimator $w(\theta)$ of $\theta$ is defined as $$\mathcal E(w(\theta)) = \frac{1/\mathcal I(\theta)}{\operatorname{Var}[w(\theta)]},$$ where $\mathcal I(\theta)$ is the Fisher information. So you need to compute the variance of each estimator.
I leave it as an exercise to show that for your distribution, the Fisher information is:
$$\mathcal I(\alpha) = \frac{n}{\alpha^2}. \tag{1}$$
The exact variance of the MLE is:
$$\operatorname{Var}[\hat \alpha] = \frac{(n \alpha)^2}{(n-1)^2 (n-2)}, \quad n > 2, \alpha > 1. \tag{2}$$
The asymptotic variance of the method of moments estimator via the CLT and the delta method is:
$$\operatorname{Var}[\tilde \alpha] = \frac{\alpha (\alpha-1)^2 ((n-1)\alpha - 2n)}{n^2 (\alpha-2)^2}. \tag{3}$$