= Solution
For $H_0=E_2B_2^\dagger B_2$, the eigenspaces of energies $0$ and $E_2$ are each two-dimensional because $n_1$ is free. <Degenerate perturbation theory> requires diagonalizing $E_1B_1^\dagger B_1$ within each space. It is already diagonal in the product <basis>, with <eigenvalues> $0,E_1$. Therefore the first-order <eigenvectors> remain
$$
|00\rangle,|10\rangle,|01\rangle,|11\rangle,
$$
and their energies are $0,E_1,E_2,E_2+E_1$, respectively. These are the exact answers, so all higher corrections vanish.
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