Prove that $\frac{\mathbb{C}[x,y]}{\langle x^2-y^2-1\rangle}$ is an integral domain such that all the nonzero prime ideals are maximal. 
Let $R=\frac{\mathbb{C}[x,y]}{\langle x^2-y^2-1 \rangle}$.
Prove that
(1) $R$ is an integral domain.
(2) Any nonzero prime ideal of $R$ is maximal.

My idea:
(1) The first part is easy. Since $x^2-y^2-1\in \mathbb{C}[y][x]$ is Eisenstein with respect to the prime $y+i$, it is irreducible. Hence the ideal generated by it is a prime ideal.
(2) I am struggling with this part. I have one approach (not sure if correct).
Approach : Let $I=\langle x^2-y^2-1 \rangle$. Then $R=\mathbb{C}[x,y]/I$.  Any ideal of $R$ is of the form $J/I$ where $J\subseteq \mathbb{C}[x,y]$ is an ideal containing $I$. If we can show that any prime ideal $J$ containing $I$ is maximal, then we are done. How to prove this?
Is there any other way to prove the second part?
Edit: I am trying to avoid Hilbert Nullstellensatz and the notion of dimension of a ring, as these concepts are not there in the syllabus of the exam from which I found this question.
 A: Solution 1. The second part is also very simple. Indeed, recall that if $k=\overline{k}$ is an algebraically closed field every prime ideal of $k[X,Y]$ is of the form $\langle X-x,Y-y\rangle$ or of the form $(f(X,Y))$ with $f(X,Y)\in k[X,Y]$ irreducible. If you take $k[X,Y]/I$ for some ideal $I$, we have that the set of the prime ideals of $k[X,Y]/I$ is simply the set of the primes in $k[X,Y]$ of the primes which contain $I$. Now, since the polynomial $X^2-Y^2-1$ is irreducible, the unique primes that contain it are those generated by at least two elements, i.e. the maximals in $\mathbb{C}[X,Y]$, that remain maximal when we think them in $\mathbb{C}[X,Y]/(X^2-Y^2-1)$. 
Solution 2. The ring $R=\mathbb{C}[X,Y]/(X^2-Y^2-1)$ is isomorphic to $\mathbb{C}[X,Y]/(XY-1)$ via the map which takes $X+Y$ and $X-Y$ to $X$ and $Y$, respectively. Moreover, the ring $A=\mathbb{C}[X,Y]/(XY-1)$ is isomorphic to the localization of $\mathbb{C}[X]$ at $X$, i.e. $A\simeq B:=\mathbb{C}[X,1/X]$, via the map which send $X$ to $X$ and $Y$ to $1/X$. Now the primes of $B$ are those of the form $(X-x)$, with $x\in\mathbb{C}^\times$, therefore we have done since every $(X-x)\subset B$ is maximal and $R\simeq B$.
