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I will now show that there are not any solutions with $q=2$, $g\geq 3$ or $q=3$, $g \geq 2$. This will be a bit messy. Let $q$ be $2$ or $3$. Since $q$ isn't square, $\sqrt{q}$ and $- \sqrt{q}$ must have the same multiplicity as eigenvalues; call this multiplicity $k$. Let the other eigenvalues be $\sqrt{q} e^{\pm i \theta_j}$ for $1 \leq j \leq g-k$. Then the equations $\#X(\mathbb{F}_q) = \#J(\mathbb{F}_q)=1$ become $$(q+1-2 \sqrt{q})^k(q+1+2 \sqrt{q})^k \prod_{j=1}^{g-k} (q+1-2 \sqrt{q} \cos \theta_j) = q+1 - 2 \sqrt{q} \sum_{j=1}^{g-k} \cos \theta_j = 1$$ or $$\prod_{j=1}^{g-k} (q+1-2 \sqrt{q} \cos \theta_j) = \frac{1}{(q-1)^{2k}} \qquad \sum_{j=1}^{g-k} \cos \theta_j = \tfrac{\sqrt{q}}{2}.$$ We will consider minimizing $\prod_{j=1}^{g-k} (q+1-2 \sqrt{q} \cos \theta_j)$ subject to the constraint $\sum_{j=1}^{g-k} \cos \theta_j = \tfrac{\sqrt{q}}{2}$ and will show that, almost always the minimum is too high.

Put $c_j = \cos \theta_j$, so $-1 \leq c_j \leq 1$ and $\sum c_j = \tfrac{\sqrt{q}}{2}$; note that this is a convex polytope. The function $\log \prod_{j=1}^{g-k} (q+1-2 \sqrt{q} \cos \theta_j)$ is concave, so it is minimized at a vertex of the polytope. At a vertex of the polytope, all but one of the $c_j$ must be $\pm 1$, and the remaining $c_j$ is determined by the constraint $\sum c_j = \tfrac{\sqrt{q}}{2}$.

So the minimum either has $g-k = 2m+1$, with $$c_1=c_2=\cdots=c_m=1,\ c_{m+1}=c_{m+2}=\cdots=c_{2m}=-1,\ c_{2m+1}=\tfrac{\sqrt{q}}{2}$$ or else the minimum has $g-k=2m+2$ with $$c_1=c_2=\cdots=c_{m+1}=1,\ c_{m+2}=c_{m+2}=\cdots=c_{2m+1}=-1, \ c_{2m+2}=\tfrac{\sqrt{q}}{2}-1.$$

The resulting values for the product are $(q-1)^{2m}$ and $(q-1)^{2m} (2 q^{3/2} - 3q +1)$. In all cases, this is larger than $\tfrac{1}{(q-1)^{2k}}$, except for the first case with $m=k=0$, $q=2$ or $3$, $g=1$ and the second case with $m=k=0$, $q=2$ and $g=2$ and one more case that I missed the first time around. That last case is $q=2$, $g=3$, in which case we have equality at the vertices of the polytope (which is a hexagon). These vertices are the permutations of $(1, -1, 1/\sqrt{2})$, giving eigenvalues $(\sqrt{2}, \sqrt{2}, -\sqrt{2}, - \sqrt{2}, 1+i, 1-i)$. However, such a curve would have $-3$ points over $\mathbb{F}_4$, which is impossible, so we are safe.

In the genus one cases, equality occurs at the sole point of the polytope, with $c_1=\tfrac{\sqrt{q}}{2}$. In the genus two case, the polytope $c_1+c_2 = \tfrac{\sqrt{2}}{2}$ is a line segment. The value of the product at the ends of the line segment is $<1$ and at the midpoint is $>1$, so equality occurs at a pair of symmetric points in the interior of the line segment, namely at $(\tfrac{1 \pm \sqrt{3}}{2 \sqrt{2}},\tfrac{1 \mp \sqrt{3}}{2 \sqrt{2}})$. These give the eigenvalues listed above.

I attempted to prove that the curves above were the only examples, but my proof had a gap. Fortunately, pregunton discovered a paper of Mercuri and Stirpe which lists all curves with $J(\mathbb{F}_q)$ trivial (both with $X(\mathbb{F}_q)$ singleton, like we want, and with $X(\mathbb{F}_q)$ empty). The curves listed above are cases (i), (ii), (vi) and (vii) on their list; I believe all the other curves on their list have $X(\mathbb{F}_q)$ empty.

Here is what I can salvage from my computation, which I still like:

Now let $q$ be $2$ or $3$. We observe that $\sqrt{q}$ must have even multiplicity as an eigenvalue of Frobenius. Proof: We have $\#J(\mathbb{F}_q) = \prod (1-\lambda_j)$. Group together the terms with complex conjguate eigenvalues; then every factor except the one coming from $1-\sqrt{q}$ is a positive real, so $(1-\sqrt{q})$ must be raised to an even power as well. Using $\#J(\mathbb{F}_q) = \prod (\lambda_j-1)$, we see that $-\sqrt{q}$ likewise has even multiplicity.

Therefore, we can write the eigenvalues of Frobenius as $\sqrt{q} e^{\pm i \theta_j}$ for $1\leq \theta_j \leq g$, remembering to take both signs in the exponent even when $\theta_j$ is $0$ or $\pi$. Let $c_j = \cos \theta_j$.

The equation $X(\mathbb{F}_q)=1$ translates to $$q-2 \sqrt{q} \sum \cos \theta_j +1 =1 \ \implies \ \sum c_j = \tfrac{\sqrt{q}}{2} \quad (\ast)$$ The equation $J(\mathbb{F}_q)=1$ translates to $$\prod (q+1-2 \sqrt{q} c_j) = 1 \quad (\dagger).$$ We note that $(\ast)$ is a convex polytope whose vertices are $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2})$ for $g$ odd and $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2}-1)$ for $g$ even.

We will try to show that $\prod (q+1-2 \sqrt{q} c_j) \geq 1$ everywhere on this polytope. Note that $\log \prod (q+1-2 \sqrt{q} c_j)$ is concave, so it is enough to check the inequality at the vertices.

We compute that the value at the vertices is $$\begin{array}{c|cc} & g=2k+1 & g=2k+2 \\ \hline q=2 & 1 & (3-2 \sqrt{2}) (3 - 2 \sqrt{2} (\tfrac{\sqrt{2}}{2}-1)) \approx 0.657 \\ q=3 & 4^k & 4^k (4-2 \sqrt{3}) (4 - 2 \sqrt{3} (\tfrac{\sqrt{3}}{2}-1)) \approx 2.39 \times 4^k \\ \end{array}.$$

We see that the only solution with $q=3$ is $g=1$ with $c_1 = \tfrac{\sqrt{3}}{2}$.

If $q=2$ and $g=2k+1$, then $(\dagger)$ only occurs at the vertices of the polytope, corresponding to the eigenvalue sequence $\sqrt{2}$ (multiplicity $2k$), $-\sqrt{2}$ (multiplicity $2k$) and $1 \pm i$. However, this would lead to a curve with $1-4k$ points over $\mathbb{F}_4$, so this can only happen for $k=0$ (and thus $g=1$.)

However, I am unclear as to how to deal with the case of $q=2$, $g$ even. When $g=2$, the polytope is just a line segment and $(\dagger)$ occurs at two symmetrically placed points of the line segment, corresponding to $(c_1, c_2) = ( \tfrac{1 \pm \sqrt{3}}{\sqrt{2}}, \tfrac{1 \mp \sqrt{3}}{\sqrt{2}})$. However, for larger $k$, $(\dagger)$ occurs on a little $2k$-dimensional manifold in the corners of the polytope, and it isn't clear to me how to rule out more solutions here.

The resulting values for the product are $(q-1)^{2m}$ and $(q-1)^{2m} (2 q^{3/2} - 3q +1)$. In all cases, this is larger than $\tfrac{1}{(q-1)^{2k}}$, except for the first case with $m=k=0$, $q=2$ or $3$, $g=1$ and the second case with $m=k=0$, $q=2$ and $g=2$.

The resulting values for the product are $(q-1)^{2m}$ and $(q-1)^{2m} (2 q^{3/2} - 3q +1)$. In all cases, this is larger than $\tfrac{1}{(q-1)^{2k}}$, except for the first case with $m=k=0$, $q=2$ or $3$, $g=1$ and the second case with $m=k=0$, $q=2$ and $g=2$ and one more case that I missed the first time around. That last case is $q=2$, $g=3$, in which case we have equality at the vertices of the polytope (which is a hexagon). These vertices are the permutations of $(1, -1, 1/\sqrt{2})$, giving eigenvalues $(\sqrt{2}, \sqrt{2}, -\sqrt{2}, - \sqrt{2}, 1+i, 1-i)$. However, such a curve would have $-3$ points over $\mathbb{F}_4$, which is impossible, so we are safe.

So the minimum either has $g-k = 2m+1$, with $$c_1=c_2=\cdots=c_m=1,\ c_{m+1}=c_{m+2}=\cdots=c_{2m}=-1,\ c_{2m+1}=\tfrac{\sqrt{q}}{2}$$ or else the minimum has $g-k=2m+2$ with $$c_1=c_2=\cdots=c_{m}=1,\ c_{m+1}=c_{m+2}=\cdots=c_{2m+1}=-1, \ c_{2m+2}=\tfrac{\sqrt{q}}{2}-1.$$

So the minimum either has $g-k = 2m+1$, with $$c_1=c_2=\cdots=c_m=1,\ c_{m+1}=c_{m+2}=\cdots=c_{2m}=-1,\ c_{2m+1}=\tfrac{\sqrt{q}}{2}$$ or else the minimum has $g-k=2m+2$ with $$c_1=c_2=\cdots=c_{m+1}=1,\ c_{m+2}=c_{m+2}=\cdots=c_{2m+1}=-1, \ c_{2m+2}=\tfrac{\sqrt{q}}{2}-1.$$