Finding the maximal ideals of the quotient of a polynomial ring by an ideal

Let the field $k$ be algebraically closed, let $f(X) \in k[X]$ be a separable polynomial of degree at least $2$, let $$B = \frac{k[Y,X]}{(Y^2 - f(X))}$$ and write $y,x$ for the images in $B$ of $Y$ and $X$, respectively.

The Nullstellensatz tells me that the maximal ideals of $k[Y,X]$ are of the form $(X-\alpha, Y-\beta)$ for $\alpha,\beta\in K$. But what do the maximal ideals of $B$ look like?

I can write $B = k[y,x]$, but this is not isomorphic to the polynomial ring $k[Y,X]$ because $y^2 = f(x)$. Thus, I don't think I can directly apply the Nullstellensatz to $B$. Is there another way?

Recall the ideal correspondence for quotient rings: The ideals of a ring of the form $R/I$ are ideals of $R$ which contain $I$. This is just like the subgroup correspondence for quotient groups.
So you are right to start off by noting the maximal ideals of $k[x,y]$ are of the form $(x-a, y-b)$ by the Nullstellensatz. Now any maximal ideal $m$ of your ring $B$ must also be of the form $(x-a,y-b)$ but it must also contain the defining ideal $(y^2 - f(x))$. Therefore, we must have $y^2 - f(x)$ in $(x-a,y-b)$. One can work out at this point that this means that this is equivalent to the condition $b^2 = f(a)$.
Its even easier to see this geometrically! In particular, by denoting $V(y^2 - f(x))$ the vanishing locus in $k^2$, i.e., the set of points $(a,b)$ such that $b^2 = f(a)$, and recalling that the Nullstellensatz gives a correspondence between such general points $(a,b)$ in $k^2$ and maximal ideals $(x-a,y-b)$, we see that $(y^2 - f(x)) \subseteq (x-a,y-b)$ if and only if the point $(a,b)$ lies on the curve $V(y^2 -f(x))$. This recovers again the condition $b^2 = f(a)$.