A general result states that an $R$-module $M$ is flat if and only if $I\otimes_R M \simeq IM$ for all ideals $I\subset R$. However, there is something I don't understand.
Let $R = \mathbb{Z}$, and let $M = \mathbb{Z}/m$. Then, for any ideal $I = n\mathbb{Z}$, it seems to me that $$IM = n\mathbb{Z} \cdot \mathbb{Z}/m\simeq n (\mathbb{Z}/m)$$ and that $$I \otimes_\mathbb{Z} M = n\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/m = \mathbb{Z} \otimes_\mathbb{Z} n(\mathbb{Z}/m) \simeq n(\mathbb{Z}/m).$$ What is my mistake? I mean, I'm aware of that $\mathbb{Z}/m$ is not flat...
What you've written makes no sense unless $n$ is a divisor of $m$. The correct statements are $$IM=n\mathbb{Z}\cdot \mathbb{Z}/m\mathbb{Z}=\gcd(n,m)\mathbb{Z}/m\mathbb{Z}$$ and because $I\cong \mathbb{Z}$, $$I\otimes_\mathbb{Z}M\cong\mathbb{Z}\otimes_\mathbb{Z}M\cong M=\mathbb{Z}/m\mathbb{Z}$$ which are only isomorphic when $\gcd(n,m)=1$. Since for any $m>1$ there is some $n$ with $\gcd(n,m)>1$, the module $M$ is not flat.
(I also suggest you get in the habit of writing out quotient $\mathbb{Z}/m\mathbb{Z}$ in full, with the second $\mathbb{Z}$.)