Prove that if $m\mid z$ implies that $n \mid z$, then $n\mid m$ I am trying to prove the following divisibility fact: 

For any $z \in \mathbb{Z}$, suppose $m \mid z$ implies that $n \mid z$. Then $n \mid m$.

The intuition here is not too difficult for me to grasp. If, for example, some $z$ always has a factor of $m$ implies that it always has a factor of $n$, that would suggest we can take a factor of $n$ out of $m$. I am having some difficult proving this fact rigorously, though. 
A hint on how to proceed would be greatly appreciated. Is contradiction the appropriate approach? 
 A: Set $z=m$. Clearly $m \mid m$. We have:
$$n \mid z  \implies n \mid m$$
Done! 
BTW we can substitute $z=m$ since the condition is:
$$m \mid z \implies n \mid z$$
for any $z$.
A: It's clearer in terms of sets of multiples, where divides = contains by $\,(\color{#c00}{3\!\iff\! 2})\,$ below. The utility of this view becomes clear when one studies (principal) ideals or (cyclic) groups.
Lemma $ $ TFAE for $\,m,n\in\Bbb Z$
$(1)\ \ \, n\:\!|\:\!z\,\Leftarrow\, m\:\!|\:\!z,\ $ for all $\,z\in\Bbb Z$
$\!\left.\begin{align}&\color{#c00}{(2)}\ \ \ \ n\Bbb Z\,\supseteq\, m\Bbb Z\\[.5em]
&\color{#c00}{(3)}\ \ \ \ \ \ \  n\,\mid\, m\end{align}\,\right\}\ $ [divides = contains]
Proof $\ \ (1\Rightarrow 2)\,\ \ z\in m\Bbb Z\,\Rightarrow\,m\mid z\Rightarrow\,n\mid z\,\Rightarrow\, z\in n\Bbb Z$
$(2\Rightarrow 3)^{\phantom{|^|}}\ \ m\in m\Bbb Z\subseteq n\Bbb Z\,\Rightarrow\, m\in m\Bbb Z\,\Rightarrow\,m = nz\,\Rightarrow\,n\mid m.\,$
$(3\Rightarrow 1)^{\phantom{|^|}}\ \ n\mid m\mid z\,\Rightarrow\, n\mid z\ $ by transitivity of "divides".
