I'm supposed to use the fact that if $A \in M_{m,n}(R)$ with $m<n$, then the columns of $A$ are linearly dependent. This is true since $A$ represents a map $f: R^n \to R^m$ and the columns of $A$ span the image of $f$. Thus, the number of columns is greater than the rank of $R^n$, which is $n$.
Now, if $M=\langle m_1, \dots, m_k \rangle$, then there is an epimorphism $R^k \to M$.
How could we continue using the above exercise?
If $M=\langle m_1,\ldots,m_k\rangle$, then we can write any $m\in M$ as $$m=a_1m_1+\cdots+a_km_k$$ Suppose we have a set $n_1,\ldots,n_{l}\in M$ with $l>k$, and let $n_i=a_{i,1}m_1+\cdots+a_{i,k}m_k$. Then we have a matrix $$\begin{pmatrix}a_{1,1}&\cdots&a_{l,1}\\\vdots & \ddots & \vdots \\a_{1,k} & \cdots & a_{l,k}\end{pmatrix}$$ Now, since we have more columns than rows, we must have a dependence, say $$b_1\begin{pmatrix}a_{1,1}\\\vdots\\a_{1,k}\end{pmatrix}+\cdots+b_l\begin{pmatrix}a_{l,1}\\\vdots\\a_{l,k}\end{pmatrix}=0$$ Then $$b_1n_1+\cdots+b_ln_l=0$$ and so the $n_i$ cannot be linearly independent.