If $P$ is a prime ideal and $xy\in P$, then either $x\in P$ or $y\in P$, and so by induction, if $x^n\in P$, then $x\in P$. Therefore, any prime ideal containing $I$ actually contains $J=\langle 6, x-2, y \rangle$, and actually must contain one (and only one) of $J_2=\langle 2, x-2, y \rangle$ or $J_3=\langle 3, x-2, y \rangle$ (it cannot contain both, as an ideal which contains both $2$ and $3$ is trivial, and at least some definitions of prime ideal require the ideal to be proper). However, both of these ideals are maximal, as their quotients are the fields with $2$ or $3$ elements, respectively.
To see the isomorphism, we note the following fact: If $R$ is a ring, then $R[x]/(x-a)\cong R$. This follows from the first isomorphism theorem by taking the map $R[x]\to R$ sending $x$ to $a$. A slight extension of this yields that $\mathbb Z[x,y]/\langle 6, x-2, y \rangle\cong \mathbb Z[x]/\langle 6, x-2\rangle \cong \mathbb Z/\langle 6\rangle$. Actually, this gives us another perspective, as we can phrase the original problem as quotients of $\mathbb Z/\langle 6\rangle$ which are domains/fields.
For your second question, note that if an ideal contains both $y$ and $p(x,y)$ as generators, then you can replace $p(x,y)$ with $p(x,0)$ (which is the remainder upon dividing $p$ by $y$). Doing this first with $y$, then $x-2$, you have replaced $p$ with a constant polynomial.
The hard part of this is showing that the only commutative rings with additive group $(\mathbb{Z}/2\mathbb{Z})^3$ are the six you mentioned; the other additive groups have at most two generators and I imagine you've completed that part.
Here I'll give two proofs. The second proof is much shorter than the first.
First Proof
To this end, I will call the generators of $(\mathbb{Z}/2\mathbb{Z})^3$ $1,e_2$ and $e_3$ (WLOG the multiplicative identity is a generator, since any nonzero element can be a generator). And before fixing our perspective on $e_2$ and $e_3,$ I will split it into two cases: rings with a subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ and rings with no such thing.
Let's do the second case first. Notice that there are only three multiplications to fix to determine the entire ring: $e_2^2, e_2e_3,$ and $e_3^2.$ Now since we said that there is no subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ it must be true that $1, e_2,$ and $e_2^2$ generate the whole group; thus WLOG $e_3=e_2^2.$
Since now $e_3^2=e_2(e_2e_3),$ there is in fact only one multiplication to set, $e_2e_3.$ Of the eight possibilities, four immediately give errors: $e_2e_3$ cannot be equal to $0, e_2, e_3,$ or $e_2+e_3+1,$ or we would obtain that $\{0,1,e_3,e_3+1\}$ is a subring. Thus we are left with:
- $e_2e_3=1$
- $e_2e_3=e_2+1$
- $e_2e_3=e_3+1$
- $e_2e_3=e_2+e_3.$
But in fact the first and fourth of these give errors as well, since $\{0,1,e_2+e_3,e_2+e_3+1\}$ is a subring in these cases.
Another way to say the second and third of these is:
- $e_2^3-e_2-1=0$
- $e_2^3-e_2^2-1=0.$
Both of these clearly define $\mathbb{F}_8.$
Now we consider the case where there is a subring isomorphic to $\mathbb{F}_4, \mathbb{F}_2[x]/(x^2),$ or $\mathbb{F}_2^2.$ In each case let us say that that subring is $\{0,1,x,x+1\},$ and the other four elements are $\{y,y+1,x+y,x+y+1\}.$ Either $xy$ is to be found in the former subring or it isn't. In the first case by choosing $y$ appropriately we may have $xy=0$ in the case of $\mathbb{F}_4,$ and we may have either $xy=0$ or $xy=1$ in the cases of $\mathbb{F}_2[x]/(x^2)$ and $\mathbb{F}_2^2.$ Since the value of $y^2$ hasn't been set yet, this leaves us a priori with 40 possibilities. But in case $xy=0,$ we must have that $y^2$ is annihilated by $x,$ which is 2/8 for $\mathbb{F}_4$ and 4/8 for $\mathbb{F}_2[x]/(x^2), \mathbb{F}_2^2.$ And in case $xy=1,$ we must have $xy^2=y,$ which has been rendered impossible. So there are only 10 possibilities, in each of which $xy=0$:
- $\mathbb{F}_4,~y^2=0$
- $\mathbb{F}_4,~y^2=y$
- $\mathbb{F}_2[x]/(x^2),~y^2=0$
- $\mathbb{F}_2[x]/(x^2),~y^2=x$
- $\mathbb{F}_2[x]/(x^2),~y^2=y$
- $\mathbb{F}_2[x]/(x^2),~y^2=x+y$
- $\mathbb{F}_2^2,~y^2=0$
- $\mathbb{F}_2^2,~y^2=x+1$
- $\mathbb{F}_2^2,~y^2=y$
- $\mathbb{F}_2^2,~y^2=x+y+1.$
But the sixth case repeats the fifth under the replacement $z=x+y,$ and the eighth repeats the seventh under the replacement $z=x+y+1.$
The first ring is $\mathbb{F}_2[z]/(z^3),$ where the generator of the $\mathbb{F}_4$ subring is $z+1$ and $y=z^2.$ The second is clearly $\mathbb{F}_4\times\mathbb{F}_2.$ The third is $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$ verbatim, the fifth is $\mathbb{F}_2[x]/(x^2)\times\mathbb{F}_2,$ and the fourth is $\mathbb{F}_2[y]/(y^3).$ The seventh repeats the fifth under $x\leftrightarrow y,$ the ninth is clearly $\mathbb{F}_2^3,$ and the last is $\mathbb{F}_4\times\mathbb{F}_2,$ where $(\omega,0)=y$ and $(0,1)=x.$
If $xy$ is one of the "upper four," then all possibilities for the value of $xy$ in the cases of the subrings $\mathbb{F}_4$ or $\mathbb{F}_2[1+i]$ lead to contradictions with the associativity of $x^2y$ (here $x=\omega,1+i$ resp.)
In the case of $\mathbb{F}_2^2,$ if $xy$ is one of the "upper four" and $y^2$ is one of the "lower four," then we can say WLOG either $y^2=0$ or $y^2=x.$ This makes a priori 8 possibilities. But only the following two do not lead to contradictions:
- $\mathbb{F}_2[e],~y^2=0,~xy=y$
- $\mathbb{F}_2[e],~y^2=x,~xy=y.$
(I have written $\mathbb{F}_2^2$ using an idempotent $e=x.$)
The former is $\mathbb{F}_2[i]\times\mathbb{F}_2,$ where $y=(1+i,0)$ and $x=(0,1),$ and the latter is the former under the transformation $z=x+y.$
If $xy$ and $y^2$ are in the "upper four," we still have a priori 16 cases to consider. But in fact we must have either $xy=y$ or $xy=x+y+1$ for the associativity of $x^2y,$ thus there are only eight.
- $\mathbb{F}_2^2,~xy=y,~y^2=y$
- $\mathbb{F}_2^2,~xy=y,~y^2=y+1$
- $\mathbb{F}_2^2,~xy=y,~y^2=x+y$
- $\mathbb{F}_2^2,~xy=y,~y^2=x+y+1.$
- $\mathbb{F}_2^2,~xy=x+y+1,~y^2=y$
- $\mathbb{F}_2^2,~xy=x+y+1,~y^2=y+1$
- $\mathbb{F}_2^2,~xy=x+y+1,~y^2=x+y$
- $\mathbb{F}_2^2,~xy=x+y+1,~y^2=x+y+1.$
($x$ is an idempotent nonunit of $\mathbb{F}_2^2$ in each case.)
But the second and fourth violate the associativity of $xy^2.$ I'll leave it to you to determine which of the first, third, and fifth through eighth are valid rings, and which rings in your list they correspond to.
Second Proof
This begins as a converse to what you noted. We have either a field, a ring with a unique nonzero maximal ideal, or a nonlocal ring. In the case of a field we know it's $\mathbb{F}_8.$ In the case of a nonlocal ring we have Theorem 3.1.4 on page 40 of Finite Commutative Rings and Their Applications by Bini and Flamini:
A finite, commutative ring with identity, R, can be expressed as a direct sum of local rings. This decomposition is unique up to permutation of direct summands.
(It's a good read!) and so this leaves only the local case.
The maximal ideal $\mathfrak{m}$ has either two or four elements, and we'll rule out $|\mathfrak{m}|=2$ at the end.
In the case of four elements, let's call those elements $0,x,y,x+y.$ If $\mathfrak{m}^2=0$, we have $x^2=xy=y^2=0,$ so we get $\mathbb{F}_2[x,y]/(x^2,xy,y^2).$
If $\mathfrak{m}^2\neq0,$ then we can't have both $x^2=0$ and $y^2=0,$ or else $(x+y)^2=0,$ and thus $(xy)^2=0$ leads to a contradiction in any of the three choices $xy=x,~xy=y,~xy=x+y$ by multiplying through by $x$ or $y.$ But $xy\in\mathfrak{m}$ then implies $xy=0,$ which means $\mathfrak{m}^2=0,$ contradiction. So WLOG $y=x^2,$ our eight elements are the at-most-quadratic polynomials in $x,$ and we need only determine which of the four elements of $\mathfrak{m}$ is $x^3.$ The choice $x^3=0$ leads to $\mathbb{F}_2[x]/(x^3).$ The choices $x^3=x$ and $x^3=x^2$ lead to $x+1$ being uninvertible, which is a contradiction since any element outside of $\mathfrak{m}$ must be invertible. And the choice $x^3=x^2+x$ leads to $x^2+x+1$ being uninvertible.
Finally, if $|\mathfrak{m}|=2,$ then $R/\mathfrak{m}\cong\mathbb{F}_4$ and $R$ has six units. Take a unit $x$ other than 1. Then $\bar x=\omega$ by an isomorphism, $\bar x^2=\omega+1,$ and $\bar x^3=1,$ and $x,x^2,x^3$ are all units in $R.$ Say $\mathfrak{m}=\{0,y\}.$ We can't have $y=x+1$ because they go to different things in the quotient, and so we have a unique way to write all eight elements of $R.$ Now $xy\in\mathfrak{m}$ implies $xy=0$ or $xy=y.$ But note too that $(x+1)y=0$ or $(x+1)y=y.$ Only two of the four possibilities do not lead directly to a contradiction and they are symmetric, so we may say $xy=0.$ But then if $x^2=x+1,$ multiplying through by $y$ gives us $y=0,$ and if $x^2=x+y+1$ then multiplying through by $y$ gives us $y^2=y.$ These are the only two possibilities since $\bar x^2=\overline{x+1},$ the former is a contradiction since $y\neq0,$ and the latter gives us $R\cong\mathbb{F}_4\times\mathbb{F}_2,$ which is a contradiction since we said $R$ is local.
Best Answer
It is not hard to show that the set of zerodivisors of the ring $\mathbb C[x,y]$ with $xy=0$ is the union of the prime ideals $(x)$ and $(y)$. (One knows this without any proof since the set of zerodivisors is the union of associated prime ideals and it's easily seen that $(XY)=(X)\cap(Y)$.) Then $S$, the set of non-zerodivisors of $\mathbb C[x,y]$, is nothing but $\mathbb C[x,y]-(x)\cup(y)$. In particular, this shows that the total ring of fractions $S^{-1}\mathbb C[x,y]$ is a semilocal ring having exactly two maximal ideals. Now all it's done.