Consider a finite field $\mathbb{F}\_q$ such that $q \equiv 1 \pmod3$ (i.e., $\omega \mathrel{:=} \sqrt[3]{1} \in \mathbb{F}\_q$ for $\omega \neq 1$) and an element $b \in (\mathbb{F}\_q^\*)^2 \setminus (\mathbb{F}\_q^\*)^3$ (a quadratic residue and cubic non-residue in $\mathbb{F}\_q$). Further, I am interested in the elliptic $\mathbb{F}\_q$-surface (i.e., elliptic $\mathbb{F}\_q(t)$-curve)
$$
E\!: \begin{cases}
y\_1^2 - b = b(y\_0^2 - b)t^3,\\
y\_2^2 - b = b^2(y\_0^2 - b)t^3
\end{cases} \quad \subset \quad \mathbb{A}^4\_{(y\_0,y\_1,y\_2,t)}.
$$
It is readily checked that $E$ has a Weierstrass form $W\!: y^2 = x^3 + a\_4x + a\_6$ with
\begin{align\*}
a\_4 \mathrel{:=}{} & -3b^4s^4 + 3b^2(b + 1)s^3 - 3(b^2 - b + 1)s^2, \\ 
a\_6 \mathrel{:=}{} & -2b^6s^6 + 3b^4(b + 1)s^5 + 3b^2(b^2 - 4b + 1)s^4 - (2b^3 - 3b^2 - 3b + 2)s^3,
\end{align\*}
where $s \mathrel{:=} t^3$. Note that $W$ has the infinite order $\mathbb{F}\_q$-section (i.e., $\mathbb{F}\_q(t)$-point)
$$
x \mathrel{:=} \big(b(2bs - 1) - (3bs - 2) \big)s, \qquad y \mathrel{:=} 3\sqrt{b}(b - 1)(2\omega + 1)s^2(bs - 1).
$$
Besides, the Mordell–Weil rank of $W$ over $\overline{\mathbb{F}\_q}(s)$ (and hence over $\mathbb{F}\_q(s)$) is equal to $1$. This immediately follows if we look at the degenerate fibers of $W$ as a rational elliptic surface. What about the Mordell–Weil rank of $W$ over $\overline{\mathbb{F}\_q}(t)$ and $\mathbb{F}\_q(t)$?


Let us identify $W$ with the corresponding Kodaira–Néron model, which is clearly a $K3$ surface. Again, analyzing the degenerate cases of $W$, we see that
$\operatorname{rk}W\big( \mathbb{F}\_q(t) \big) \leqslant \operatorname{rk}W\big( \overline{\mathbb{F}\_q}(t) \big) \leqslant 5$ (even $\leqslant 3$ if $W$ is non-supersingular), because, as is known, the Picard $\overline{\mathbb{F}\_q}$-number $\rho(W) \leqslant 22$ (resp., $\leqslant 20$ if $W$ is non-supersingular).


It seems that $W$ is a singular $K3$ surface (i.e., $\rho(W) \mathrel{:=} 20$) and $\operatorname{rk}W\big( \mathbb{F}\_q(t) \big) = 1$. At least, for some small $q$ and $b$ (satisfying the restrictions) I checked these conjectures by means of the CAS Magma.



Mordell–Weil rank of some elliptic $K3$ surface
Let $E\_0$ be the elliptic curve $y^2 = x^3 + 1$,
and choose $\beta$ in $k = {\bf F}\_q$ so that $b = \beta^2$.
Then $W = W\_b$ has $\rho=18$ unless the elliptic curve
$$
E\_\beta : Y^2 = X^3 + \beta \, (3X + (\beta+1)^2)^2
$$
is isogenous to $E\_0$, in which case $\rho(W)$ is
$20$ ("singular") or $22$ ("supersingular")
depending on whether $E\_0$ is ordinary or supersingular,
that is, depending on whether the characteristic of $k$ is
$1 \bmod 3$ or $-1 \bmod 3$.
(This does not depend on the choice of square root $\beta$ of $b$,
because the curve $E\_\beta$ and $E\_{-\beta}$ are $3$-isogenous over $\bar k$.)


The recipe for $\rho(W\_b)$ holds whether or not $b$ is a cube,
and over any field of characteristic not $2$ or $3$
(the supersingular case does not occur in characteristic zero).


In particular if $b \in {\cal B} := \{-1, 1/4, 4, 4/5, 5/4, 27/28, 28/27\}$
then $E\_0$ is isogenous with $E$, so $W$ is supersingular
in characteristic $-1 \bmod 3$, and singular otherwise.
The fact that ${\cal B}$ is symmetrical under $b \leftrightarrow 1/b$
reflects an isomorphism between $W\_b$ and $W\_{1/b}$.


The elliptic fibration over the $t$-line is $y^2 = {\rm cubic}(x)$
with a cubic that factors completely, so instead of narrow Weierstrass form
we write it as
$$
 y^2 = x \, (x - b^2 t^4 + t) \, (x - b^2 t^4 + b t).
$$
This works over any field where $6 \neq 0$,
and becomes equivalent with the OP's formula once we scale $x,y$
by powers of $\sqrt{-3}$. The isomorphism between
$E\_b$ and $E\_{1/b}$ is $(x,y,t,b) \leftrightarrow (x,y,bt,1/b)$.
The nontorsion section is $(x,y) = (b^2 t^4, b^{3/2} t^3)$,
which has canonical height $3/2$; this section,
together with the reducible fibers and $2$-torsion sections,
gives a sublattice of ${\rm NS}(W\_b)$ of rank $18$ and discriminant $-144$.


We next move to another elliptic fibration with parameter $u$ where
$$
x = u t (b t^3 - 1).
$$
This yields an even nicer equation
$$
Y^2 = X^3 + b (u-1)^2 u^3 (u-b)^4.
$$
Its singular fibers at $u=0,\infty$ (type I${}\_0^\*$), $u=1$ (type IV),
and $u=b$ (type IV${}^\*$) account for the same sublattice of ${\rm NS}(W\_b)$.


This new elliptic fibration tells us that $W\_b$ is the quotient of
$E\_0 \times C\_b$ by a $6$-cycle, where $E\_0: y^2 = x^3 + 1$ as above
and $C\_b$ is the curve
$$
C\_b: v^6 = b (u-1)^2 u^3 (u-b)^4.
$$
The $6$-cycle is generated by the product of
the automorphisms of $E\_0$ on $C\_b$ that take
$(x,y)$ to $(\zeta^2 x, \zeta^3 y)$ and $(u,v)$ to $(u, \zeta^{-1} v)$
where $\zeta$ is a primitive sixth root of unity (so $\zeta^3 = -1$).


Now $C\_b$ has genus $2$, with hyperelliptic involution
$(u,v) \leftrightarrow (u,-v)$ because the quotient by this involution is
the rational curve $w^3 = b (u-1)^2 u^3 (u-b)^4$ (with $w=v^2$).
An explicit parametrization is
$$
w = u (u-1) (u-b) z, \quad u = \frac{z^3-b^2}{z^3-b},
$$
and we find that $C\_b$ is birational with the curve
$\eta^2 = b (z^3 - b) (z^3-b^2)$. Recalling that $b = \beta^2$,
we write $z = \beta(1+T)/(1-T)$ and find the equivalent model
$$
U^2 = \beta \left(
 (\beta+1)^2 T^6 - 3(\beta^2-10\beta+1)T^4
 + 3(\beta^2+10\beta+1)T^2 - (\beta-1)^2
 \right)
$$
of $C\_b$. The quotients of $C\_b$ by the non-hyperelliptic involutions
$(T,U) \leftrightarrow (-T,U)$ and $(T,U) \leftrightarrow (-T,-U)$
then give quadratic twists of the elliptic curves $E\_{\pm\beta}$.
We conclude that the Jacobian of $C\_b$ is isogenous over $\bar k$ with
$E\_\beta \times E\_{-\beta}$, and thus with $E\_\beta^2$;
we soon recover from this our recipe for the Picard number of $W\_b$.


The list $\cal B$ of rational values of $b$ for which $W\_b$ is singular
(or supersingular in characteristic $-1 \bmod 3$) is then obtained by
setting the $j$-invariant of $E\_\beta$ equal to the rational or quadratic
$j$-invariants of the CM elliptic curves with discriminant
$-3, -2^2 3, -3^2 3, -5^2 3, -7^2 3$.


