Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented

Condition 1 implies condition 2, because $M$ is a submodule of itself and a Noetherian module has all of its submodules finitely generated.

Condition 2 implies condition 1, since $R$ is a Noetherian ring and a finitely generated module over such a ring is Noetherian.

Condition 3 implies condition 2: an exact sequence $R^m\to R^n\to M\to 0$ has its right-hand map $\beta$ surjective, because exactness at $M$ says the image of $\beta$ is the kernel of the zero map $M\to 0$, which is all of $M$; every element of $R^n$ is ${\sum }_{i=1}^n{r}_i{e}_i$, so every element of $M$ is ${\sum }_{i=1}^n{r}_i\beta (e_i)$ and $M$ is generated by the finite set $\{\beta (e_1),\dots ,\beta (e_n)\}$.

Assume condition 2 and fix a finite generating set $m_1,\dots ,m_n$ of $M$, with $n\in \mathbb{N}$. The universal property of the free module gives an $R$-module homomorphism $\beta \text{\hspace{0.17em}⁣}:R^n\to M$ with $\beta (e_i)=m_i$, and its image is the set of all finite sums ${\sum }_i{r}_i{m}_i$, which is $⟨m_1,\dots ,m_n{⟩}_R=M$; so $\beta$ is surjective.

Still assuming condition 2, the module $R^n$ is generated by the finite set $\{e_1,\dots ,e_n\}$, hence is Noetherian over the Noetherian ring $R$; therefore its submodule $K:=\ker \beta$ is finitely generated, say by $k_1,\dots ,k_m$ with $m\in \mathbb{N}$.

Still assuming condition 2, the universal property gives $\alpha \text{\hspace{0.17em}⁣}:R^m\to R^n$ with $\alpha (e_j)=k_j$, whose image is $⟨k_1,\dots ,k_m{⟩}_R=K=\ker \beta$; together with the surjectivity of $\beta$ this makes $R^m\to R^n\to M\to 0$ exact, so $M$ is finitely presented and condition 2 implies condition 3.

Steps 1.1, 1.2, 1.3 and 3.1 close the cycle: condition 1 gives condition 2, condition 2 gives condition 1 and condition 3, and condition 3 gives condition 2. The three conditions are therefore equivalent.