Is there a way to find any $\mathbb{F}_2(t)$-point on the elliptic curve $\mathcal{E}$?

Question

Consider the ordinary elliptic curves $$ E\!:y_1^2 + x_1y_1 = x_1^3 + 1,\qquad E^\prime\!: y_2^2 + x_2y_2 = x_2^3 + x_2^2 + 1 $$ over the field $\mathbb{F}_2$. They are quadratic twists to each other. I checked that the Kummer surface of $E \!\times\! E^\prime$, i.e., the quotient $E \!\times\! E^\prime/[-1]$ has the affine model $$ K\!: y^2 + x_1x_2y = (x_1x_2)^2(x_1 + x_2 + 1) + (x_1 + x_2)^2, $$ where $y := x_1y_2 + y_1x_2$.

Let $t := x_1$, $x := x_2$. It is almost obvious that the elliptic curve $K_t/\mathbb{F}_2(t)$ is reduced (over $\mathbb{F}_2(t)$) to the form $$ \mathcal{E}\!:y^2 + txy = x^3 + t(t^2 + t + 1)x^2 + t^4x. $$

Is there a way to find any $\mathbb{F}_2(t)$-point on $\mathcal{E}$ outside $$\mathcal{E}[2] = \{(0:0:1), (0:1:0)\}?$$

Answer 1

Any $\mathbb F_2(t)$-point of $K_t$ would give, upon pullback to $E \times E'$, a $\mathbb F_2(E)$-point of $E'$. Because $E$ is ordinary, $a_2(E)\neq 0$, hence $E$ is not isogenous to its quadratic twist $E'$, so any such point arises from an $\mathbb F_2$-point of $E'$. Because $E'$ has two $\mathbb F_2$-points, these are the only ones.