Example of $A subseteq k[x,y]$, $dim(A)=2$, $J$ non-maximal ideal of $k[x,y]$ and $J cap A$ a maximal ideal of $A$

After receiving interesting answers to this question (see also this question), I wish to concentrate on the following special case:
Let $A subseteq B=k[x,y]$, $k$ is a field of characteristic zero.
Assume that $dim(A)=dim(B)$ (of course, $dim(B)=2$).

Question 1: Is it possible to find
a non-maximal ideal $J$ of $B=[x,y]$ such that $J cap A$ is a maximal ideal of $A$?

The answer to question 1 is yes, with the following being an example of Youngsu:

$A=k[x,xy]$, $J=xk[x,y]=(x)_{k[x,y]}$ contracts to the maximal ideal $(x,xy)A=(x,xy)_{k[x,xy]}$.

So now I add the condition that $J$ is generated by two algebraically independent elements of $B$,
but still gets an example:
$A=k[x,y(y-1)]$, $J=xk[x,y]+y(y-1)k[x,y]=(x,y(y-1))_{k[x,y]}$ contracts to
$xA+y(y-1)A=(x,y(y-1))_A$, which is clearly maximal in $A$. ($J$ is non-maximal in $B$ and also non-prime in $B$).

So now I ask:

Question 2: ‘When’ such $J$ (= two generated by two alg. ind. elements of $B$)
must be maximal (or at least prime) in $B$, given its contraction $J cap A$ is maximal in $A$?

Ideas for ‘When’:

(1) $A subseteq B$ is integral: Not a good idea, since in my last example $A subseteq B$ is integral,
but $J$ is not a maximal ideal of $B$ (and also not a prime ideal of $B$).

(2) $A subseteq B$ is flat: (I have not yet tried to figure out if $k[x,y(y-1)] subseteq k[x,y]$ is flat or not).
For faithfully flat extension $A subseteq k[x,y]$, we have for every ideal
$I$ of $A$: $Ik[x,y]cap A=I$.

(3) $Q(A)=Q(B)$: I do not know if the fraction fields are relevant here.

(4) $A subseteq B$ is separable: I think that in my last example the extension is not separable,
but do not see any connection between my question and being separable.

Any hints and comments are welcome!