$p$ is a positive integer and $(p)$ is a maximal ideal in the ring $(\mathbb Z, +,\cdot)$, then $p$ is a prime number I need to prove: $p$ is a positive integer and $(p)$ is a maximal ideal in the ring $(\mathbb Z, +,\cdot)$, then $p$ is a prime number. 
My attempt: 
1) $(p)$ is a maximal ideal, so it is a prime ideal.
2) Assume $p$ is not prime, then $p = ab, a,b \in \mathbb Z, a>0, b>0,a \ne 1,p$. 
3) Because $(p)$ is a prime ideal, it means that either $b=z_1p$ or $a=z_2p, z_1>0,z_2>0, z_1,z_2 \in \mathbb Z$. 
4) Assume that $b=z_1p$, then $1=az_1$ so $a= 1, z_1 =1$.
5) Assume that $a=z_2p$, then $1=bz_2$, so $b= 1, z_1 =1$, but then $a=p$.
Thus $p$ must be prime.
Is it a correct proof?

As you noticed yourself ideal $\left(p\right)$ is also a prime ideal. 
So $nm\in\left(p\right)\implies n\in\left(p\right)\vee m\in\left(p\right)$,
or equivalently $p\mid nm\implies p\mid n\vee p\mid m$. 
This combined with $p\notin\{0,1,-1\}$ makes us say that $p$ is a so called
prime element of $\mathbb{Z}$.
This combined with the positivity of $p$ makes us say that $p$ is a prime number.

edit:
Every prime element is an irreducible element. Proof:
Let $p$ be a prime element and let $p=ab$. 
Then $p\mid a\vee p\mid b$. If $p\mid a$ then $a=rp$ for some $r$. 
Then $p=ab$ leads to $p(1-rb)=0$ and consequently $rb=1$ showing that $b$ is a unit. 
Of course $p\mid b$ will lead to the conclusion that $a$ is a unit.

Applied to $\mathbb Z$ this shows that a prime number can be described as a positive integer that has exactly $2$ positive divisors: $1$ and itself.
