Which "known" ring is $K = \frac{\mathbb{Z}_5[X,Y]}{(Y-X^2, XY + Y + 2)}$ isomorphic to?

Very good.

You can prove that it's isomorphic to $\Bbb Z_5[X]/(X^3+X^2+2)$ and you already showed that polynomial has no roots in $\Bbb Z_5$.

Hence, it is irreducible (being of degree 3), so the quotient is the field of $5^3$ elements.

Note that by your remark $(Y-X^2,XY+Y+2) = (Y-X^2,X^3+X^2+2)$.

The assignment $X \to X, Y\to X^2$ defines an arrow $\tau \colon \Bbb Z_5[X,Y]/(Y-X^2) \to \Bbb Z_5[X]$. Surjectivity is clear, as for injectivity: pick $p(X,Y)$ in the kernel of $\tau$ and via the identification $\Bbb Z_5[X,Y] = \Bbb Z_5[X][Y]$ divide $p$ by $Y-X^2$. Note that even though polynomial rings in several variables need not have euclidean algorithms, we can always divide by monomials (as in this case). Hence we have

$$ p(X,Y) = (Y-X^2)q(X,Y) + r(X). $$

But then applying $\tau$ we get $r(X) = 0$ as desired.

Finally, by the third isomorphism theorem we have

$$ \Bbb Z_5[X,Y]/(Y-X^2, X^3+X+2) \simeq \Bbb Z_5[X]/(X^3+X^2+2). $$

Note that the polynomial $X^3+X^2+2$ has no roots modulo $5$ and is thus irreducible in $\Bbb Z_5$. Therefore, the quotient is a field. To compute the order, you can note for example that it is a $\Bbb Z_5$ vector space of dimension $3$ over $\Bbb Z_5$.

Hence the quotient can be characterized as the field with $5^3$ elements.