From Lang, Undergraduate Algebra:
Theorem 3.2 Suppose that $R$ is an integral ring (commutative ring without zero-divisors and such that $1\ne 0$). Then the integer n such that $\mathbb{Z}/n\mathbb{Z}$ is contained in $R$ must be $0$ or a prime number.
Note: "$\mathbb{Z}/n\mathbb{Z}$ is contained in $R$" means that $f(\mathbb{Z}/n\mathbb{Z})\subseteq R$, where $f$ is the unique ring homomorphism from $\mathbb{Z}$ to $R$ (conserving the unit).
Proof Suppose $n$ is not $0$ and is not prime. Then $n = mk$ with integers $m, k\ge 2$, and neither $m, k$ are in the kernel of the homomorphism $f:\mathbb{Z}\to R$. Hence $me\ne 0$ and $ke\ne 0$. But $(me)(ke) = mke= 0$, contradicting the hypothesis that $R$ has no divisors of $0$. Hence $n$ is prime.
Note: $e=1_R$.
My question:Why neither $m, k$ are in $\ker f$? Intuitively, if $m\in\ker f$, then $f(m)=0$ and $f(mk)=f(n)=0$, and $f(n)=0$ should be a contradiction, why? Or which is the right argument?
I think there is an implicit sentence that you forgot to "read". Since we have a canonical ringmorphism from $\mathbb{Z}$ ($1_\mathbb{Z}\mapsto 1_R$ ), it has a kernel which is a subgroup and hence of the form $n\mathbb{Z}$. In particular you get by the homomorphism theorem an injective map $$\mathbb{Z}/n\mathbb{Z}\hookrightarrow R$$ (here the implicit argument ends and the normal proof continues)
Now you know that if $n$ is not prime there exist $m,k < n$ such that $mk=n$, in particular $[m]\neq0\neq [k] \in \mathbb{Z}/n\mathbb{Z}$ but $[m][k]=[mk]=[n]=[0]=0\in \mathbb{Z}/n\mathbb{Z}$. So both of those would be zero devisors, and you are done.