I think that the correct spectrum is: $E_n = \frac{\hbar |eB|}{m}(n+1/2)$.
To solve the problem we start noticing that, defining:
$$x' := P_{1}+\frac{1}{2}eBX_{2}\quad \mbox{and}\quad p':= P_{2}-\frac{1}{2}eBX_{1}$$
we have:
$$[x',p'] = i\hbar eB I$$Assuming $eB >0$ (otherwise it is sufficient to swap the definitions of $x$ and $p$), we can redefine
$$x := |eB|^{-1/2}x'\quad \:, \quad p := |eB|^{-1/2}p'$$ so that, they both hold:
$$[x,p] = i \hbar I\qquad (1)$$
and
$$H=\frac{|eB|}{2m}(x^2+p^2)\:.\qquad (2)$$
consequently we have the standard levels of a harmonic oscillator:
$$E_n = \frac{\hbar |eB|}{m}(n+1/2)\:,\qquad (3)$$
NOTE. The answer could stop here. However something further can be said also taking into account the kind correction by nervxxx to my previous version, concerning the degeneracy of levels.
If the CCR representation (1) were irreducible we would have a harmonic oscillator with eigenspaces of dimension 1.
However this representation is not irreducible because the Hilbert space is ${\cal H}= {\cal H_1} \otimes {\cal H_2} = L^2(R) \otimes L^2(R)$ as initially assumed and there are operators commuting with $x$ and $p$.
One of them is:
$$\tilde{H} =\frac{|eB|}{2m}(\tilde{x}^2+\tilde{p}^2)\:,$$
where we have defined (using the same strategy exploited to define $x$ and $p$ but changing the internal signs)
$$\tilde{x} := |eB|^{-1/2} \left(P_{1}-\frac{1}{2}eBX_{2}\right)\quad \mbox{and}\quad \tilde{p}:= |eB|^{-1/2} \left( P_{2}+\frac{1}{2}eBX_{1}\right)$$
We conclude that in each eigenspace of $\tilde{H}$ we have a harmonic oscillator described by the restriction of $H$ to this invariant subspace.
Since the eigenvalues of $\tilde{H}$ are (countably) infinitely many we conclude that the degeneracy of each level of $H$ is infinite.
