I find the answer in papers that studies bilayer grapheme, e.g.
http://iopscience.iop.org/article/10.1088/0034-4885/76/5/056503/meta;jsessionid=6653715AE8C3DDEC60ADA7854E2EA192.c1
and I decided to write the answer to my own question. First we do minimal coupling to the magnetic field:
\begin{equation}
H[A]=\frac{(k_x+e A_x/c)^2-(k_y+e A_y/c)^2}{m}\sigma_x+\frac{(k_x+e A_x/c)(k_y+e A_y/c)+(k_y+e A_y/c)(k_x+e A_x/c)}{m}\sigma_y
\end{equation}
Notice that $k_x k_y$ should be symmetrized when replaced with canonical momentum in order to keep the Hermicity of the Hamiltonian. In Landau gauge,
\begin{equation}
A_x=-B y, A_y=0
\end{equation}
then
\begin{equation}
[\frac{(-i \partial_x-\frac{e B}{c} y)^2-(-i \partial_y)^2}{m}\sigma_x+\frac{(-i \partial_x-\frac{e B}{c})(-i \partial_y)+(-i \partial_y)(-i \partial_x-\frac{e B}{c})}{m}\sigma_y]\psi(\mathbf{r})=E_n \psi(\mathbf{r})
\end{equation}
Due to translational invariance in x-direction,
\begin{equation}
\psi(\mathbf{r})=\frac{1}{\sqrt{L}}exp[i k x]\hat{f}_n(y)
\end{equation}
one find
\begin{equation}
[\frac{(k-\frac{e B}{c} y)^2-(-i \partial_y)^2}{m}\sigma_x+\frac{(k-\frac{e B}{c})(-i \partial_y)+(-i \partial_y)(k-\frac{e B}{c})}{m}\sigma_y]\hat{f}(y)=E_n \hat{f}(y)
\end{equation}
Defining the creation and annihilation operator as
\begin{equation}
a^-=l_B \partial_y+(l_B k-\frac{e B}{c} y/l_B),
a^+=l_B \partial_y-(l_B k-\frac{e B}{c} y/l_B),
\end{equation}
where $l_B=\sqrt{\frac{c}{e B}}$ is the magnetic length. We have
\begin{equation}
\omega_c
\begin{bmatrix}
0 & {a^+}^2 \\
{a^-}^2 & 0
\end{bmatrix}
\begin{bmatrix}
f_n^+(y) \\
f_n^-(y)
\end{bmatrix}=E_n
\begin{bmatrix}
f_n^+(y) \\
f_n^-(y)
\end{bmatrix}
\end{equation}
The spectrum and eigenstates can be solved in analogy with Harmonic oscillator problem:
\begin{equation}
E_n^{\pm}=\sqrt{n(n-1)}\omega_c,n=2,3,\ldots, \hat{f}_{n,\pm}=\frac{1}{\sqrt{2}}
\begin{bmatrix}
\phi_n(y) \\
\pm\phi_{n-2}(y)
\end{bmatrix}
\end{equation}
where $\phi_n(y)$'s are eigenstates of Harmonic oscillators.