How to tell if an element of a quotient ring is a zero divisor

I am looking at Hartshorne Example III.9.8.4., p260. He says that $a$ is not a zero divisor in $k[a,x,y,z]/I$, where $$ I = (a^2(x+1) -z^2, ax(x+1)-yz, xz-ay,y^2-x^2(x+1)). $$ Is there a good way to see this?

Let me replace $a$ with $w.$ First, consider the domain $S = k[x,y]/(y^2-x^2(x+1)).$ Then $S[z,w]$ is a graded ring ($\mathbb N$-grading) generated in degree $1$ over $S$ by $\{z,w\}.$ Now $w^2(x+1)-z^2,wx(x+1)-yz,xz-wy$ are homogeneous elements of $S[z,w]$ of respective degrees $2,1,1.$ Thus the ideal generated by these elements is homogeneous, and the quotient $R = S[z,w]/(w^2(x+1)-z^2,wx(x+1)-yz,xz-wy)$ is graded. In particular, both $z,w$ still have degree $1$ in the quotient ($x,y$ are in degree $0$) and so cannot possibly be zerodivisors, since their products with any nontrivial elements must have degree at least $1.$ But $R = k[x,y,z,w]/(w^2(x+1)-z^2,wx(x+1)-yz,xz-wy,y^2-x^2(x+1))$ is exactly our original ring, so we are done.

Edit:

This argument about degrees isn't very convincing. How about the following:

The quotient ring $S = k[x,y,z] / \langle y^2 - x^2(x+1) \rangle$ is a domain, so injects into its quotient field $Q(S)$. This yields an injection $S[a] \hookrightarrow Q(S)[a]$. Now let $J \subset S[a]$ be defined by the three remaining relations. Over $Q(S)[a]$, these all reduce to $a = xz/y$, so $Q(S)[a] / JQ(S)[a] = Q(S)$. Thus there is a canonical map $S[a]/J \to Q(S)$ such that $a \mapsto xz/y$. It is clear that if $a$ is a zero-divisor, then this map must send $a$ to $0$, which is false. Hence $a$ cannot be a zero-divisor.