= Solution
Use the ordered <computational basis> $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$, with $Z|0\rangle=|0\rangle$ and $Z|1\rangle=-|1\rangle$. Write $X,Y,Z$ for the <Pauli matrices>, $I$ for the <identity operator>, $J=\omega_{12}$, $\Sigma=\omega_1+\omega_2$, and $\Delta=\omega_1-\omega_2$. As in the question, the <Hamiltonian operators> have <angular frequency> units, so the <time-evolution operator> is $e^{-iHt}$. The <Heisenberg coupling matrix for two spins> is
$$
C=X\otimes X+Y\otimes Y+Z\otimes Z
=\begin{pmatrix}
1&0&0&0\\
0&-1&2&0\\
0&2&-1&0\\
0&0&0&1
\end{pmatrix}.
$$
Consequently the full <Heisenberg model> <Hamiltonian operator> and its proposed <Ising coupling> approximation are
$$
H_{\mathrm H}=\begin{pmatrix}
(\Sigma+J)/2&0&0&0\\
0&(\Delta-J)/2&J&0\\
0&J&(-\Delta-J)/2&0\\
0&0&0&(-\Sigma+J)/2
\end{pmatrix},\qquad
H_{\mathrm I}=\operatorname{diag}\!\left(
\frac{\Sigma+J}{2},\frac{\Delta-J}{2},
\frac{-\Delta-J}{2},\frac{-\Sigma+J}{2}\right).
$$
For <nondegenerate perturbation theory>, choose $H_{\mathrm I}$ as the unperturbed <Hamiltonian operator> and
$$
V=H_{\mathrm H}-H_{\mathrm I}
=J\bigl(|01\rangle\langle10|+|10\rangle\langle01|\bigr)
$$
as the perturbation. Only $|01\rangle$ and $|10\rangle$ are coupled, and their unperturbed <energy levels> differ by $\Delta$. All first-order <energy> corrections vanish. The first-order <eigenvectors> and second-order <energy> corrections are
$$
|\widetilde{01}\rangle=|01\rangle+\frac{J}{\Delta}|10\rangle
+O\!\left(\frac{J^2}{\Delta^2}\right),\qquad
|\widetilde{10}\rangle=|10\rangle-\frac{J}{\Delta}|01\rangle
+O\!\left(\frac{J^2}{\Delta^2}\right),
$$
$$
\delta E_{01}^{(2)}=\frac{J^2}{\Delta},\qquad
\delta E_{10}^{(2)}=-\frac{J^2}{\Delta}.
$$
The states $|00\rangle,|11\rangle$ are exact <eigenvectors> throughout. Hence the <weak coupling condition for two NMR spins> is
$$
\boxed{|\omega_1-\omega_2|\gg |\omega_{12}|.}
$$
A large common <Zeeman energy> alone does not ensure this condition. Crossings involving the uncoupled parallel-spin states do not invalidate the argument: the relevant denominator is the separation of the two states connected by $V$.
As an exact check, the middle $2\times2$ <matrix> has <eigenvalues>
$$
E_\pm=-\frac J2\pm\sqrt{\frac{\Delta^2}{4}+J^2}.
$$
For $\Delta\ne0$, labeling by the limiting <computational basis> states reproduces the shifts above. The <quantum transition probability> from $|01\rangle$ to $|10\rangle$ is
$$
P_{01\to10}(t)=\frac{4J^2}{\Delta^2+4J^2}
\sin^2\!\left(\frac{t}{2}\sqrt{\Delta^2+4J^2}\right).
$$
Thus unwanted exchange has a uniformly small amplitude when $|J/\Delta|\ll1$. If accumulated phases must also be approximated over a duration $t$, neglecting the second-order shifts additionally requires $J^2t/|\Delta|\ll1$, or those shifts must be retained in an improved effective <Hamiltonian operator>.
The same result is the <secular approximation for scalar-coupled spins>. In the <interaction picture> of the <Zeeman energy> terms,
$$
V_{\mathrm{int}}(t)=J\left(e^{i\Delta t}|01\rangle\langle10|
+e^{-i\Delta t}|10\rangle\langle01|\right).
$$
These terms average rapidly on the coupling timescale $1/|J|$ when $|\Delta|\gg|J|$, while $JZ\otimes Z/2$ survives.
For fixed nuclei and electronic environments, the individual <NMR> angular frequencies scale with the static <magnetic field> $B_0$: schematically $\omega_i=\kappa_i B_0$, where $\kappa_i$ includes the <gyromagnetic ratio> and magnetic shielding. The <NMR scalar coupling> $J$ in <angular frequency> units is approximately independent of $B_0$. Therefore a nonzero separation $|\Delta|=|\kappa_1-\kappa_2|B_0$ increases with <magnetic field>, improving the approximation. In particular, a fixed difference of <chemical shifts> gives a larger separation in hertz at higher <magnetic field>. If the two resonances remain exactly equal, increasing their common frequency does \b[not] suppress their exchange.
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