My false proof : Set $A = \mathbb{Z}[\sqrt{-3}]$ and $B = \mathbb{Z} [\frac{1+ \sqrt{-3}}{2}]$. Then as a $A$-module,
\begin{align*} B^{\oplus 2} &\cong B \otimes_{\mathbb{Z}} A \\ &\cong \mathbb{Z}[X]/(X^2 + X + 1) \otimes_{\mathbb{Z}} A \\ &\cong A[X]/(X^2 + X + 1) \\ &\cong A^{\oplus 2}. \end{align*}
Thus $B$ is a direct summand of free $A$-module. In particular, $B$ is a faithfully flat $A$-module. Indeed, let $N$ be a nonzero $A$-module. Then we get
\begin{align*} 0 &\neq (A \otimes_A N)^{\oplus 2} \\ &\cong A^{\oplus 2} \otimes_A N \\ &\cong B^{\oplus 2} \otimes_A N \\ &\cong (B \otimes_A N)^{\oplus 2} \end{align*}
and $B \otimes_A N \neq 0$.
I show another proof of faithfully flatness of $B$. We know $B$ is a projective $A$-module, so it is sufficient to show that for any maximal ideal $\mathfrak{m} \subset A$ we get $\mathfrak{m} B \neq B$. Now assume there is a $\mathfrak{m} \subset A$ such that $\mathfrak{m} B = B$. Then
$$ 2B \subset 2(\mathfrak{m} B) \subset \mathfrak{m} ( 2B) \subset \mathfrak{m} $$
So $2 \in \mathfrak{m}$. Since $A/2A \cong F_2[X]/(X+1)^2$, $\mathfrak{m}$ must be eaqual to $(2, 1 + \sqrt{-3})$. However $(2, 1 + \sqrt{-3})B = 2B \neq B$, which is a contradiction.
Question: I know the argument is incorrect. What is the first mistake of my argument?
The fact that $B$ is a direct summand of a free $A$-module only implies that $B$ is projective and flat. It does not in general imply that $B$ is faithfully flat.