If $x, y \ge 0$, we can use a 4-term AM-GM to obtain
$$x^4 + y^4 + 18 = x^4 + y^4 + 9 + 9 \geq 4 (x^4 \cdot y^4 \cdot 9 \cdot 9)^{1/4} = 12xy.$$
We get equality if and only if all four terms are equal, i.e. if and only if $x = y = \sqrt{3}$.
If one of the two numbers $x, y$ is non-positive and the other non-negative, the RHS of the target inequality is non-positive whereas the LHS is at least 18, so the inequality still holds, but with no possibility of equality.
Finally, if $x, y < 0$, we replace $(x, y)$ by $(-x, -y)$ which leaves both LHS and RHS unchanged, so the inequality holds by what we have shown already, and equality iff $x = y = -\sqrt{3}$.
In summary, we have proved the inequality and shown that equality occurs if and only if $x$ and $y$ are both equal to $\sqrt{3}$ or both equal to $-\sqrt{3}$.

$$\begin{aligned} 6(x - y)^2 &= 6x^2 + 6y^2 - 12xy, \\ (x^2 - 3)^2 &= x^4 - 6x^2 + 9 \quad \text{and} \\ (y^2 - 3)^2 &= y^4 - 6y^2 + 9, \end{aligned}$$
we have
$$x^4 + y^4 + 18 - 12xy = 6(x - y)^2 + (x^2 - 3)^2 + (y^2 - 3)^2 \geq 0$$
with equality iff $x = y = \pm\sqrt{3}$.