Let $I$ be a proper ideal in $R=\mathbb{C}[x,y]$, generated by two elements $f_1,f_2$, $I=\langle f_1,f_2 \rangle$.
I wonder if the following claim is true or false:
Claim: If for every $c_1,c_2,d,e \in \mathbb{C}$, each of the ideals $\langle f_1-c_1,f_2-c_2,x-d \rangle$ and $\langle f_1-c_1,f_2-c_2,y-e \rangle$ is either maximal or $R$, then there exist $a,b \in \mathbb{C}$ such that $I_{a,b}=\langle f_1-a,f_2-b \rangle$ is a maximal ideal of $R$.
Examples: (1) $f_1=x(x+1), f_2=y$, so $I_{a,b}=\langle x(x+1)-a,y-b \rangle$ is not maximal, for every $a,b \in \mathbb{C}$.
In this case, take $c_1=c_2=e=0$ and get that $\langle x(x+1)-0, y-0, y-0 \rangle=\langle x(x+1), y \rangle$ is not maximal and not equals $R$.
(2) $f_1=x+y, f_2=y^2+y$, so $I=\langle x+y,y^2+y \rangle$ is not maximal, it has two common zeros: $(0,0)$ and $(1,-1)$. It is clear that all $\langle x+y-c_1,y^2+y-c_2,x-d \rangle$ and $\langle x+y-c_1,y^2+y-c_2,y-e \rangle$ are maximal or $R$. Here, taking $a=0,b=-\frac{1}{4}$ seems to make $I$ maximal: $I_{0,-\frac{1}{4}}=\langle x+y,y^2+y+\frac{1}{4}\rangle=\langle x+y,(y+\frac{1}{2})^2 \rangle$; if I am not wrong it is maximal, having one common zero $(\frac{1}{2},-\frac{1}{2})$.
Remarks: (1) The one-dimensional case says: If for every $c \in \mathbb{C}$, $I=\langle f-c \rangle$ (adding no generator) is either maximal or $\mathbb{C}[x]$, then there exists $a \in \mathbb{C}$ such that $I=\langle f-a \rangle$ is maximal; here it follows that $f$ is linear (and we can take any $a \in \mathbb{C}$), since if $f=\tilde{c} \in \mathbb{C}$, then taking the scalar $\tilde{c}$ would yield the zero ideal $\langle \tilde{c}-\tilde{c} \rangle=\langle 0 \rangle$ which is not maximal and not $\mathbb{C}[x]$.
(2) I wonder if there exists a known algebraic geometry result which would solve my question.
(3) This is a relevant question.
Any comments are welcome; thank you!
EDIT: In view of the nice comment, I would like to add one of the following conditions:
Condition 1: For every $\alpha,\beta \in \mathbb{C}$, $I_{\alpha,\beta}=\langle f_1-\alpha,f_2-\beta \rangle$ is a prime ideal.
Condition 2: For every $\alpha,\beta \in \mathbb{C}$, $I_{\alpha,\beta}=\langle f_1-\alpha,f_2-\beta \rangle$ is a radical ideal.