Let $k$ be a field of characteristic $0$ and $R = k[[x_1, \dots, x_n]]$. Suppose that $M$ is a faithful, finitely generated $R$-module and $\mathfrak{a} < R$ is an ideal such that $\mathfrak{a} M = \mathfrak{m} M$, where $\mathfrak{m} = (x_1, \dots, x_n)$. Is it true that $\mathfrak{a} = \mathfrak{m}$?

I know that $\sqrt{a} = \mathfrak{m}$ and the conclusion is false if $\mathfrak{m}$ is not maximal. For example $(x_1^2, x_2^2)(x_1, x_2) = (x_1^2, x_1 x_2, x_2^2) (x_1, x_2)$ in the ring $k[[x_1, x_2]]$. However, I have not been able to find a counter-example with $\mathfrak{m}$ maximal.