Assuming the Axiom of Choice, minimal generators over a local ring are exactly residue-field bases

If the images of $x_1,\dots ,x_r$ span $M\mathrm{/}\mathfrak{m}M$, then [L1] gives that $x_1,\dots ,x_r$ generate $M$.

Suppose the images of $x_1,\dots ,x_r$ are linearly dependent over $k$. Then there are $a_1,\dots ,a_r\in R$, not all in $\mathfrak{m}$, with ${\sum }_i{a}_i{x}_i\in \mathfrak{m}M$. Choose $j$ with $a_j\mathrm{\notin }\mathfrak{m}$. Since $R$ is local, the ideal $\mathfrak{m}+(a_j)$ properly contains $\mathfrak{m}$, so by maximality it is all of $R$; choose $b\in R$ and $c\in \mathfrak{m}$ with $1=b{a}_j+c$. Multiplying the relation by $b$ shows $x_j$ lies in the submodule generated by the other $x_i$ together with $\mathfrak{m}M$. Therefore the other $r-1$ elements generate $M\mathrm{/}\mathfrak{m}M$, so [L1] makes them generate $M$. Thus the original generating set was not minimal.

Conversely, if $x_1,\dots ,x_r$ generate $M$ but are not minimal, then some $x_j$ lies in the submodule generated by the other $x_i$. Passing to $M\mathrm{/}\mathfrak{m}M$ shows that the image of $x_j$ lies in the $k$-span of the other images, so the images are linearly dependent.

Therefore $x_1,\dots ,x_r$ is a minimal generating set of $M$ if and only if its images form a $k$-basis of $M\mathrm{/}\mathfrak{m}M$. Any two minimal generating sets give two bases of the same $k$-vector space, so they have the same cardinality.