Why $M$ is not the zero module?

Question

I want to prove the following question:

A module is simple if it is not the zero module and it has no proper nonzero submodule.

Let $M$ be an $R-$module. Show that the following conditions are equivalent.

1. $M$ is a simple $R-$module.

2. There exists a maximal ideal $\mathfrak{m} \in R$ such that $M \cong_R R/ \mathfrak{m}.$

I managed to prove $1 \implies 2.$ But for $2 \implies 1,$ here is my trial:

Assume that there exists a maximal ideal $\mathfrak{m} \in R$ such that $M \cong_R R/ \mathfrak{m}.$ We want to show that $M$ is a simple $R-$module.

Since $\mathfrak{m} \in R$ is a maximal ideal, then $R/ \mathfrak{m}$ is a field. Hence the only submodules (ideals) of $R/\mathfrak{m}$ are $0$ and itself. Since $M \cong_R R/ \mathfrak{m},$ then the only submodules (ideals) of $M$ are $0$ and itself and so $M$ is simple as required.

My questions are:

1- In the definition of a simple module, it must be a non-zero module. How can I show that in my proof of $2 \implies 1$?

2- Is my proof correct (excluding the part of showing that $M \neq 0$)?

Could anyone help me in answering those questions please?