Let $d \not \in \{0, \pm 1\}$ be a square free integer. Let $\omega_1, \omega_2 \in O_d$ be a $\mathbb{Z}$-basis of the ring $O_d$ of algebraic integers of $\mathbb{Q}_\sqrt{d}$. Consider the 2 by 2 matrix $M$ whose $(i,j)$-entry is given by Tr$(\omega_i \cdot \omega_j)$. Prove that the determinant det($M$) is independent of the choice of a $\mathbb{Z}$-basis {$\omega_1, \omega_2$}.
I have initially thought that the basis can only be {$1, \sqrt{d}$} for $d\in 2+4\mathbb{Z}$ or $3+4\mathbb{Z}$ and {$1, \frac{1+\sqrt{d}}{2}$} for $d\in 1+4\mathbb{Z}$, but realised that the basis can be in the form of $a+b\sqrt{d}$ for $a, b \in \mathbb{Z}$. Upon doing some research, apparently a $\mathbb{Z}$-basis is also a power basis in the form of $\{1,d,\dots, d^{n-1}\}$, which doesn't look right as I am only taking two elements as basis. So i turned back to what I did at first, but brute force didn't get me the answer I want.
How do I proceed from here?