All-plus data have no finite Bernoulli-spin field MLE

The two statistic values are $-1$ and $+1$, so their affine hull is $\mathbb{R}$, the family is minimal, the convex hull is $[-1,1]$, and the sample mean is $1$, which is not in its interior $(-1,1)$. Hence [F2] already implies there is no finite MLE.

Let $S(h)=e^h+e^{-h}$. Both exponential terms and hence $S(h)$ are positive by [F7]. For the fixed all-plus IID sample the product likelihood is $L(h)={\prod }_{i=1}^n{\pi }_h(+1)=(e^h\mathrm{/}S(h){)}^n>0$ by [A1, F1]. Applying the positive-product law [F8] repeatedly to these $n\ge 1$ factors, then its quotient law and the inverse identity [F7], gives $\log L(h)=n(\log (e^h)-\log S(h))=n(h-\log S(h))=\mathrm{\ell }(h)$. Since $\log$ is strictly increasing, this has exactly the same maximizers as $L$. By [F3] and [F5], the two exponential terms have derivatives $e^h$ and $-e^{-h}$; [F6] therefore gives $S^{\mathrm{\prime }}(h)=e^h-e^{-h}$. The identity function has derivative $1$ directly from its difference quotient. Applying the logarithm derivative [F4] and chain rule [F5] to this derived log likelihood gives ${\mathrm{\ell }}^{\mathrm{\prime }}(h)=n(1-\frac{e^h-e^{-h}}{e^h+e^{-h}})=\frac{2n{e}^{-h}}{e^h+e^{-h}}>0$ for every finite $h$. The derived formula is differentiable on all of $\mathbb{R}$, hence continuous there: its increment is its convergent difference quotient times the real increment. For any $h_1<h_2$, [F9] applied on $[h_1,h_2]$ gives $\mathrm{\ell }(h_2)-\mathrm{\ell }(h_1)={\mathrm{\ell }}^{\mathrm{\prime }}(c)(h_2-h_1)>0$ for some $c\in (h_1,h_2)$. Thus $\mathrm{\ell }$ is strictly increasing; by [F8] the likelihood also increases. In particular each finite field has a larger likelihood at a greater field, and no finite field maximizes it.