Just going on the first sentence - I would first mod out by $(x-1)$, which has the effect of setting $x =1$ (x becomes just another symbol for 1 so we forget about it). Now the quotient is (isomorphic) to the ring $Q[y]/(y^2 + 2)$. Since $(y^2 + 2)$ is irreducible over $Q$, the ideal it generates is maximal, hence the quotient is in fact a field. But since the quotient $Q[y]/(y^2 + 2)$ is isomorphic to $Q[x,y] / (x - 1, y^2 + 2)$, the latter ring is also a field, hence your original ideal is a maximal ideal.
One can conclude $f$ irreducible hence $(f)$ maximal because $Q[y]$ is a principal ideal domain. This follows from $Q[y]$ being a polynomial ring over a field and hence a place where one can do polynomial long division, so given any ideal an element with minimal degree generates it. If there was a larger ideal $I = (f)$ containing $(y^2 + 2)$, then $f$ would be a proper divisor of $(y^2 + 2)$, and hence $(f) = (1)$.