I know that because $x^2+2$ is irreducible in $\mathbb{F}_5[x]$, then the quotient ring is a field. I'm pretty sure that the order of this quotient ring is $5*5*5 -1 =124 = 2*2*31$, because we have $5$ choices for the coefficients on $x^0,x^1$, and $x^2$, and then we have to ignore the zero element.
However, if $x$ has order $8$, then wouldn't the fact that $8 \nmid 124$ contradict Lagrange's Theorem?
On
$-2$ is not a quadratic residue $\!\!\pmod{5}$, hence $x^2+2$ is irreducible over $\mathbb{F}_5$ and $\mathbb{F}_5[x]/(x^2+2)$ is a finite field isomorphic to $\mathbb{F}_{5^2}$. The multiplicative part of a finite field is a cyclic group, hence every element of $\mathbb{F}_5[x]/(x^2+2)$ has an order that is a divisor of $24=25-1$. In particular, $x^2=-2$ implies $x^4=4$ and $x^8=1$, so the order of $x$ is exactly $8$.
You can easily verify that the order of the given element is 8: just compute all powers and check that the first one that yields 1 is the 8th power.
On the other hand, you're right when you say that the polynomial in the quotient is irreducible (there are no square roots of -2 in $\Bbb F_5$), but since the degree of this polynomial is 2, the quotient field would have $5^2 = 25$ elements, among which $25 - 1 = 24$ are invertible. Notice that since $8\mid 24$, there is no contradiction at all.