= Solution
Let $H|n\rangle=E_n|n\rangle$, with nondegenerate eigenvalues, and expand the perturbed state as $|n\rangle+\lambda|n^{(1)}\rangle+\cdots$, choosing $\langle n|n^{(1)}\rangle=0$. The first-order eigenvalue equation gives $E_n^{(1)}=V_{nn}$ and, after projection onto $m\ne n$, $\langle m|n^{(1)}\rangle=V_{mn}/(E_n-E_m)$. Projecting the second-order equation onto $n$ then gives the <second-order nondegenerate perturbation theory> result
$$
\boxed{E_n(\lambda)=E_n+\lambda V_{nn}+\lambda^2\sum_{m\ne n}\frac{|V_{mn}|^2}{E_n-E_m}+O(\lambda^3).}
$$
For <angular momentum>, $j=0,1/2,1,\ldots$, $m=-j,-j+1,\ldots,j$, and $J^2|jm\rangle=j(j+1)|jm\rangle$, $J_3|jm\rangle=m|jm\rangle$ in units $\hbar=1$.
Take $H_0=-\gamma B_3J_3$ with $\gamma B_3\ne0$ and perturbation $-\gamma B_1J_1$. The diagonal matrix element of $J_1=(J_++J_-)/2$ vanishes. Only $m\pm1$ contribute at second order. Their squared matrix elements are $[j(j+1)-m(m\pm1)]/4$, and their unperturbed energy denominators are respectively $\gamma B_3$ and $-\gamma B_3$. Thus
$$
\Delta E_m^{(2)}=\frac{\gamma B_1^2}{4B_3}\{[j(j+1)-m(m+1)]-[j(j+1)-m(m-1)]\}=-\frac{\gamma mB_1^2}{2B_3}.
$$
Hence $\boxed{E_m=-\gamma B_3m-\gamma mB_1^2/(2B_3)+O(B_1^4)}$. Rotating the quantization axis to the field gives exact energies $-\gamma m_B\sqrt{B_3^2+B_1^2}$. Labeling each branch continuously from its original $m$ gives $E_m=-\gamma mB_3\sqrt{1+B_1^2/B_3^2}$, whose expansion agrees, including negative $B_3$. If $B_3=0$, the nondegenerate starting point fails and this expansion cannot be used.
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