Consider the equation $x^2 + 2y^2 = 33^{33}$. I wish to determine the number of integers pairs $(x_{0}, y_{0})$ that satisfy this equation. My idea is to work in the ring $\mathbb{Z[\sqrt-2]}$; recall that this is a UFD since its an Euclidean Domain with map $\sigma: x+y\sqrt{-2} \to x^2 + 2y^2$. Since $33 = 11*3$ and since $11 = (3+\sqrt-2)*(3-\sqrt-2)$ and $3 = (1 + \sqrt-2)*(1 - \sqrt{-2})$, we can rewrite the Diophantine equation as:
$(x_{0}+y_{0}\sqrt-2)*(x_{0} - y_{0}\sqrt-2) = (1+\sqrt{-2})^{33}*(1 - \sqrt-2)^{33}*(3+\sqrt{-2})^{33}*(3 - \sqrt{-2})^{33}$.
Now what? I see that the elements $1 \pm\sqrt{-2}, 3 \pm \sqrt{-2}$ are prime (at least I hope they are), but I do not know what to do after this. Any suggestions, or solutions?
You are almost there! $1 \pm \sqrt{-2}$ and $3 \pm \sqrt{-2}$ are prime (they have prime norm), and they are also coprime to each other. By unique factorization, $x_0 + y_0 \sqrt{-2}$ must be a factor of the RHS, so it must factor as
$$x_0 + y_0 \sqrt{-2} = \pm (1 + \sqrt{-2})^A \cdot (1 - \sqrt{-2})^a \cdot (3 + \sqrt{-2})^B \cdot(3 - \sqrt{-2})^b.$$
Conjugating both sides,
$$x_0 - y_0 \sqrt{-2} = \pm (1 + \sqrt{-2})^a \cdot (1 - \sqrt{-2})^A \cdot (3 + \sqrt{-2})^b \cdot (3 - \sqrt{-2})^B.$$
Taking the product, we find that $A+a = B + b = 33$. Taking into account the $\pm $ sign, this gives
$$2 \times 34 \times 34 = 2312$$
possible values of $x_0 + y_0 \sqrt{-2}$ and hence the same number of pairs $(x_0,y_0)$. (One should also note that your argument is reversible --- if $x_0 + y_0 \sqrt{-2}$ is of this form, then $x^2_0 + 2 y^2_0 = 33^{33}$ and we do get a solution.