 Solution

At the maximal ideal $\mathfrak{m}=(2,X)$, the images of $4,2X,X^2$ form a three-dimensional basis of $I/\mathfrak{m}I$ over ${\mathbb{F}}_2$, so two generators are impossible. More generally $(2^{n-1},2^{n-2}X,\dots ,2X^{n-2},X^{n-1})$ needs exactly $n$ generators by the same argument.