(Note: $Z[i] = \{a + bi\ |\ a,b\in Z \}$)
This is what I have so far.
Proof:
If $I$ is a prime ideal of $Z[i]$ then $Z[i]/I$ must also be an integral domain.
Now (I think this next step is right, I'm not sure though), $$ Z[i]/I = \{a+bi + \langle 2 + 2i \rangle\ | a,b,\in Z \}. $$
So, let $a=b=2$, then we can see that $$ (2+2i) \cdot \langle 2 + 2i \rangle = \langle 4 - 4 \rangle = 0 $$
Thus, $Z[i]/I$ has a zero-divisor. Thus, $Z[i]/I$ is not an integral domain which means that $I$ is not a prime ideal of $Z[i]$.
$\square$
Now if my proof is right, am I right to think that $(2+2i) \cdot \langle 2 + 2i \rangle $ represents the following: $$ (2+2i) \cdot \langle 2 + 2i \rangle = (2+2i) \cdot \{2 +2i^2, 4 + 4i^2, 6 +6i^2, 8 + 8i^2, \ldots \} = \{0, 0, 0, 0, \ldots\} = 0? $$
Your proof is wrong: when you let $a = b = 2$ what you get is the element $2 + 2i + I$ of the quotient. You seem to then multiply this by an element of the original ring, instead of multiplying two elements of the quotient together. Moreover, your multiplication is wrong, because you've forgotten the cross terms, and anyway, the element you gave doesn't count as a zero divisor because it is in fact equal to zero.
It sounds like you need to revise quotients a bit.
An intuition for quotient rings
A quotient is basically what you do when you want to consider some ring with some elements "glued together". Say you're interested in what happens if you add the equation "$7 = 1$" to the integers. Well, by standard ring theory laws, this means that $6 = 0$ as well, and that $12 = 0$, and other things besides. What have you ended up with?
Let's try more general. Let $R$ be a ring, and $x,y$ be distinct elements. We want to set $x = y$ and see what happens. Well, if $x = y$, then $x - y = 0$, so setting two elements equal is just the same as setting another element to zero.
So let $R$ be a ring, and $z$ an element of $R$ that we want to "make zero". Then what? Well, then any multiple of $z$ must also be made zero. Suppose now that instead of just one, we want to make a collection of elements of $R$ all zero. Then clearly any sum of them is zero as well. Hmm... closed under sums and multiples, sound familiar? We conclude that to make a set $Z \subset R$ all equal to zero, we must set the entire ideal generated by $Z$ equal to zero.
So in fact we start off with setting some elements equal to each other, and realise that what we're really doing is setting some ideal equal to zero. Now, if we do that, and $z$ is in the ideal $Z$, then $z + 1$ is going to end up equal to $1$, as well. So all the elements "$a$ plus some element of the ideal $Z$" end up the same. Let's formalise this element as $a + Z$.
What's interesting about these is that because addition is an abelian group and multiplication is associative, commutative, and distributes over addition, we have things like $a + Z + b + Z = a + b + Z$ and $(a + Z)(b + Z) = ab + Z$, so in fact these elements of the form $a + Z$ themselves form a ring. This is the quotient ring: it is what you get when you start with a ring and decide you want to make some of its elements equal.
Now let's get back to your answer. If you're seeking zero divisors in the quotient ring $R/I$, you are looking for $r,s \in R$ nonzero such that $(r + I)(s + I) = 0 + I$. Now, since $(r + I)(s + I) = rs + I$, what you are looking for is $r,s \in R$ such that $r + I \not= 0 + I$ and $s + I \not= 0 + I$, but $rs + I = 0 + I$.
Let's go back to the previous interpretation. First, $0 + I$ is just "zero plus some element of $I$", which is just "some element of $I$", so $0 + I = I$ (which makes sense). Secondly, when we ask if $r + I = s + I$, we are asking if "r plus some element of $I$" is the same as "s plus some element of $I$", i.e. if there exist $a,b \in I$ such that $r + a = s + b$. But then $r - s = b - a$, so $r - s \in I$. The converse holds, too, so two elements of the quotient ring are equal if the difference of their representatives is in the ideal.
Right. So. $r + I \not= 0 + I$ is the same as $r \not\in I$, and $rs + I = 0 + I$ is the same as $rs \in I$. So the condition that $R/I$ is not an integral domain is exactly the condition that there are $r \not\in I$ and $s \not\in I$ such that $rs \in I$. We have proven:
Theorem: $R/I$ is an integral domain $\iff$ for all $a,b \in R$, $ab \in I \implies a \in I\text{ or }b \in I$.
In either case we call $I$ a prime ideal. Here's why:
Theorem: A principal ideal $\langle p \rangle$ is prime if and only if $p$ is a prime element of your ring.
Proof: The crucial observation is that $\langle p \rangle = \{x : p \mid x \}$.
With that in mind, it should be clear that $p \mid ab \implies p \mid a \text{ or }p \mid b$ is equivalent to $ab \in \langle p \rangle \implies a \in \langle p \rangle \text{ or } b \in \langle p \rangle$.
So in order to solve your problem, just show that $2 + 2i$ is not prime.