= Solution
For Neumann boundaries the correct <eigenfunctions> include zero indices:
$$
\phi_{n_x,n_y}=\cos(n_x\pi x/L_x)\cos(n_y\pi y/L_y),\qquad n_x,n_y\in\{0,1,2,\ldots\}.
$$
Their squared <wave number> is $q=(n_x\pi/L_x)^2+(n_y\pi/L_y)^2$. The constant pair $(0,0)$ is reaction-stable. Instability requires $q_-<q<q_+$, using the band derived above. If both side lengths are large, the wave-number spacings are small enough that admissible modes can enter the band in either direction, including mixed modes. This does not mean the single $(1,1)$ mode necessarily stays unstable as lengths tend to infinity: its <wave number> eventually tends to zero below the band. Rather, increasingly fine wave-number spacing permits some suitable higher integers.
<Thin-domain Turing mode selection> applies if $L_y<\pi/\sqrt{q_+}$, every mode with $n_y\geq1$ has $q\geq(\pi/L_y)^2>q_+$ and is stable. A sufficiently large $L_x$ can nevertheless admit an integer $n_x\geq1$ with $q_-<(n_x\pi/L_x)^2<q_+$ and $n_y=0$. Hence
$$
\boxed{\hbox{a sufficiently thin domain suppresses transverse variation but can retain longitudinal instability}.}
$$
The printed range starting both indices at one excludes these purely longitudinal modes and cannot imply the requested conclusion. It also omits the constant mode. Further, each allowed spatial <eigenfunction> generally has two temporal <eigenvalue> branches of $J-q\mathcal D$, with coefficients along their <eigenvectors>; a single unrestricted scalar $\lambda(q)$ is shorthand for those branches rather than a complete general expansion.
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