Show that the ring $A := \mathbb{C}[x, y]/(x^2 - y^2 - 1)$ is an integral domain. Further show that every nonzero prime ideal in A is maximal.
I proved that $A$ is an integral domain by showing that $x^2 - y^2 - 1$ is an irreducible in the UFD $\mathbb{C}[x, y]$ and hence a prime. To prove that every nonzero prime ideal in $A$ is maximal I need to prove either that:
i) A is a PID, so in a sense I need to be able to express $x$ in terms of $y$ (or vice versa) but I only know that $x^2 = y^2 + 1$ in $A$.
ii)If $I$ is a prime ideal in $A$ then every non zero element in $A/I$ is invertible.
More generally (than my suggestion in the comments): suppose that $A$ over $B$ is an extension such that every $a\in A$ satisfies a equation of the form
$$a^k+b_1 a^{k-1}+\cdots+ b_{k-1}a +b_k =0,\tag {*}$$ with $b_j \in B$, ($k$ and $b_j$ depend on $a$ of course).
(Here, $B = {\mathbb C}[y]$.)
If $A$ (and so $B$) are domains, and (*) is of minimal degree for $a\not=0$, then $b_k\not =0$.
Now, if $0\not=a\in P$ ($P$ prime), we have that $b_k \in P$. Therefore the prime $Q = B \cap P \not = 0$.
If you know that $Q$ is maximal in $B$ (as you do here!), one ends up with an extension $A/P$ over $B/Q$, where the latter is a field. To ease notation, replace $A$ with $A/P$, and $B$ with $B/Q$. The aim is to show that $A$ is a field. Using your second strategy:
Look at the equation $(*)$. Assume (once more) that the equation is of minimal degree for $a\not =0$. Rearranging, one has that $$ a ( a^{k-1} + \cdots + b_{k-1}) = - b_{k}.$$ Again, $-b_k$ is non-zero, so has a multiplicative inverse (because after relabeling, $B$ is a field). Dividing out, one gets: $$ a \cdot (-b_k^{-1})( a^{k-1} + \cdots + b_{k-1}) = 1.$$ Therefore $a$ is invertible.