You can observe that the elliptic curve $E_2: y^2=x^3+(t^3+1)^2$ is a generic fibre of a rational elliptic surface. Over the algebraically closed field $k$ the group of $k(t)$ points on the curve has rank equal to 2 (by the Shioda-Tate formula) and from the classification of possible groups of $k(t)$-rational points by Oguiso-Shioda (case 39 because we have three places of reduction type $IV$ away from characteristic 3) the group of $k(t)$-rational points modulo torsion has a lattice structure $A_2^{*}$. So we are looking at the points of height $1/2$ and this can be found only when one hits two singular points in the Weierstrass equation at places of bad reduction. So you obtain points of the form

$P_i=((4^{1/3}\cdot \zeta_3^i)(1 - t + t^2), \sqrt{-3}(-1 + t)(1 - t + t^2)$ for i=0,1,2

We have the relation $P_0+P_1+P_2=0$ and th points $P_1,P_2$ span a lattice of type $A_2^*$, hence are free generators.

Now you have to apply the map $t\mapsto t^2$ and on the elliptic curve $y^2=x^3+(t^6+1)^2$ you obtain two linearly independent points $P_1'$ and $P_2'$. Notice that $P_1'-P_2'$ is the negative of the point found by Noam Elkies.

The points are defined over $\overline{\mathbb{Q}}(t)$. By reduction they will become naturally points over $\overline{\mathbb{F}}_{p}(t)$ and can provide points defined over $\mathbb{F}_{p}(t)$.

With a trace formula and point count (which can be done in Magma as pointed out by Jeremy Rouse in his answer) you can check that in several characteristics the rank over $\mathbb{F}_{p}(t)$ can be higher than $2$. This follows from Tate conjecture and is unconditional because the given elliptic curve is a generic fibre of a K3 surface. For example: in characteristic $p=5,11,13,17$ the rank over $\mathbb{F}_{p}$ is equal to $4$.

**Addendum:**

One can play the same game with the elliptic curve $E_{tw}=E_{1}^{(t)}: y^2=x^3+(t+1)^2\cdot t^3$, which under the base change $t\mapsto t^6$ becomes isomorphic to the original curve. Curve $E_{tw}$ is a generic fibre of the rational elliptic surface with rank over $\overline{\mathbb{Q}}(t)$ equal to $2$ again (now the singular fibres have different reduction type: $I_{0}^{*}$, $II$ and $IV$.). This is the case $32$ in the Oguiso-Shioda table. We find the following points of required height

$Q_{i}=(\zeta_{6}^{i}(1+t)t, \sqrt{-1}t^2(1 + t))$
for $i=1,3,5$.

There is the equality $Q_{1}+Q_{3}+Q_{5}=0$ and the points $Q_{1}$ and $Q_{3}$ are lineary independent.

Now, we base change again to the elliptic curve $y^2=x^3+(t^6+1)^2$ and obtain the following set of four linearly independent points obtained via a base change

$$\{P_{1}',P_{2}', Q_{1}',Q_{3}'\}$$

The Gram matrix is a block matrix of determinant 4.

**Second addendum:**

Finally, we consider a curve $F:y^2=x^3+(t^3-3t)^2$. It is isomorphic to $y^2=x^3+(t^6+1)^2$ under the pullback by $t\mapsto t+1/t$. Curve $F$ has three places bad reduction of type $IV$. Hence we find from the Oguiso-Shioda the points of height $1/2$. Those points are

$$R_{i}=(\zeta_{6}^{i}(t^2 - 3),\sqrt{3}(-3 + t^2))$$
for $i=1,3,5$.
We have $R_{1}+R_{3}+R_{5}$ is zero and we finally obtain the set of 6 linearly independent points
$$\{P_{1}',P_{2}', Q_{1}',Q_{3}',R_{1}',R_{3}'\}$$
which have a Gram matrix of determinant $4/3$.

The calculations above for several characteristics show that we cannot have more than 6 independent points in the group $E(\overline{\mathbb{Q}}(t))$, so up to finite index we found a basis of points. From this using the Galois action on the points you can work out the exact value of the $\mathbb{F}_{p}(t)$ rank in the cases when the Tate conjecture computation matches this bound. For the primes where the rank over $\overline{\mathbb{F}}_{p}(t)$ is higher than $6$ the extra points might exist for very peculiar reasons, e.g. the Shioda-Inose structure might throw some extra divisors into the picture and the heights of those points can be very large.