Properties of Prime Number in Prime Ideal (Ring theory) We have the following theory:
Let $J$ be an ideal of $\mathbb{Z}$, then $J$ is a prime ideal if and only if $J= n\mathbb{Z}$ for some prime number $n$. 
Problem
The problem is with the proof. I understand that any ideal of $\mathbb{Z}$ must be in the form $n\mathbb{Z}$ for some integers $n$. But why 
$n|xy$ if and only if $n|x$ or $n|y$ implies that $n$ is a prime number? Frankly, whenever I see the divisor sign I fret because primes feel so frightening to me, but it appears here and there in abstract algebra. 
 A: In any commutative ring $R$, one defines $p\in R\setminus \{0\}$ to be a prime element of $R$ if it is not a unit and for all $a,b\in R$, $p|ab\implies p|a\text{ or }p|b$.
With this definition you can say:

For a nonzero $x\in R$, the principal ideal $(x)$ is a prime ideal iff $x$ is a prime element of $R$.

(This fixes the problem mentioned in the comments about the way you originally phrased it.)
This is nothing deep, it just rephrases the definition I gave above.
The case of $\mathbb Z$ is a very special one where all the ideals are principal, and the prime elements are the ones we are used to. 

Frankly, whenever I see the divisor sign I fret because primes feel so frightening to me

What? Why?  Of all abstract algebra topics, I would bet this is the most familiar and easy to grasp.
In commutative rings, divisibility is just a statement about containment between principal ideals. That is, $a|b \iff (b)\subseteq (a)$. Actually, this partial order gives the set of principal ideals of the ring the structure of a lattice order.
Rephrased with containment graphs, the definition of prime says that if:

then one of the following graphs must also hold 
So $p$ has a special property compared to other elements in this partial order.
A: Claim: We have $$''n|xy \iff n|x \quad \textrm{or} \quad n|y''$$ if and only if $n$ is prime.
Proof:
Only if: We prove by contradiction. If $n$ is not prime, say $n = ab$ with neither of $a$, $b$ equal to $\pm n$, then $n|ab$ but $n \not | a$ and $n \not | b$.
If: If $n$ is prime, then the fundamental theorem of arithmetic (i.e., thinking about prime decompositions) tells us that the given property holds. If a prime number appears in the prime decomposition of a product of two numbers, it must appear in the prime decomposition of one of the numbers. $ \qquad\square$
Really, in ring theory, this property is the definition of a prime element, and it happens that the prime numbers (along with their negatives) are exactly the prime elements of the ring $\mathbb{Z}$. The familiar definition of prime numbers actually means exactly that they (and their negatives) are the irreducible elements of the ring $\mathbb{Z}$, and, because $\mathbb{Z}$ is a UFD (which is another way of stating the fundamental theorem of arithmetic), the irreducible elements are exactly the prime elements.
If you still have questions, please comment and I will try to expand my answer.
