Here is Einsteins's criterion for general rings I am using:
Let's write $\mathbb C[x,y,z,w] \cong \mathbb C[x,y,z][w]$.
Now, the ideal $(z)$ is prime in $\mathbb C[x,y,z]$ since $\mathbb C[x,y,z]/(z) \cong \mathbb C[x,y]$ is an integral domain.
Next, consider the polynomial $xw-yz$ as a polynomial in $w$ over $\mathbb C[x,y,z]$, i.e., $xw-yz \in \mathbb C[x,y,z][w]$.
Now by Eisenstein:
$\bullet$ $x \notin (z)$
$\bullet$ $-yz \in (z)$
$\bullet$ $-yz \notin (z^2)$,
and so $xw-yz$ is not the product of polynomials of degree $<1$ in $\mathbb C[x,y,z][w]$?
Therefore, $xw-yz$ is irreducible in $\mathbb C[x,y,z][w]$. Since $\mathbb C[x,y,z][w]$ is a UFD, then $xw-yz$ is prime and thus $(xw-yz)$ is a prime ideal. Hence $\mathbb C[x,y,z,w]/(xw-yz)$ is an integral domain.
Have I made correct use of Eisenstein's criterion? Is this sufficient?
Yes, your use of Eisenstein's criterion is perfectly correct. The only problem is that you conclude that
But the above does not immediately imply that $xw-yz$ is irreducible. In fact the first sentence is already clear without Eisenstein's criterion; polynomials of degree $<1$ are constant, and so their product is also constant, so certainly not equal to $xw-yz$.
In fact, as stated the theorem does not seem directly useful for proving that $xw-yz$ is irreducible. On the other hand, since this polynomial is linear in $w$, if it is reducible then it must factor as $$xw-yz=c\cdot q(w),$$ where $c\in\Bbb{C}[x,y,z]$ and $\deg q=1$. But then $c$ divides both $x$ and $yz$, so it is a unit in $\Bbb{C}[x,y,z]$ and hence in $\Bbb{C}[x,y,z,w]$, which shows that $xw-yz$ is irreducible.
On the other hand, Eisenstein's criterion in its slightly more general form, with a very similar proof, tells you that the polynomial is irreducible if the $a_i$ together generate the whole ring $R$.