Here are the things that you can do with the modular form $f$ corresponding to an elliptic curve $E$:
(a) Determine the number of points on $E$ mod $p$ by computing $a_p(f)$ (easy
for smallish primes via modular symbols computations).
(b) Compute (perhaps with some effort) a modular parameterization of $E$,
and then, by evaluating this at Heegner points, find a point of infinite order
on a twist $E_D$ of $E$, in the cases when this twist has rank one.
(c) Compute whether or not $L(E_D,1) = 0$ for every twist $E_D$ of $E$,
via modular symbols. If you grant BSD, this tells whether or not the
twist $E_D$ has infinitely many points.
I'm not sure what other facts about $E$ you are expecting to get. What is it
you would like to know about an elliptic curve in any case? For most people,
the rank (and especially whether or not it is positive) is the main thing, and
conjecturally this is what you can get from the $L$-function of $E$, which is
essentially inaccessible without modular forms, but is highly computable once
you know $f$. (And not just for $E$, but for all its twists.)
Maybe the other thing you might like to know is Sha of $E$, but this is not
proven to be finite in general. Nevertheless, modular forms can sometimes
be used to witness non-trivial elements of Sha. (Read about the theory of
``the visible part of Sha'', by Cremona and Mazur.)
Consider the following elements in the field $H=\Bbb Q(\sqrt{-1},\sqrt{-5})$:
$$
\begin{aligned}
k &= \sqrt{-1}\ ,\\
a &= \sqrt{-5}\ ,\\
ak &=\sqrt{5} \text{ (in a manner compatible with the choices for $a,k$)}
\\[3mm]
s &=\frac 12(\sqrt{-1}+\sqrt{-5})\ .
\end{aligned}
$$
Then $s$ is an algebraic integer,
the minimal polynomial of $s$ is $F=X^4 + 3X^2 + 1$, so
$s$ is of norm one, and $1,s,s^2,s^3$ turns out to be an integral basis of $H$.
The reduction modulo two of $F$ is $f=x^4 + x^2 + 1=(x^2+x+1)^2$,
so it is natural to consider $\Bbb F_4:=\Bbb F_2[u]$,
where $u$ has minimal polynomial $(x^2+x+1)$.
There is thus a map $\Phi$ from the ring $\mathcal O_H$ to $\Bbb F_4$ defined
on the generator $s$ by $s\to u$.
In particular $\Phi(s^2)=u^2=u+1$.
The elements $a,k,ak$ have the following expressions in terms of $s$:
$a=s^3+4s$, $k=s^3 +2s$, $ak=-2s^2-3$. We compute $\Phi(2)=0$, $\Phi(1+ak)=0$,
so $\Phi$ induces a map from $\mathcal O_H$ modulo $(2,1\pm ak)$
(which corresponds to $(2,1+\sqrt 5)$ in either choice of the square root)
to $\Bbb F_4$.
We will use $\Phi$ when passing from an equation defined over $\mathcal O_H$ to one over $\Bbb F_4$. (I.e. we apply the tensor functor $-\otimes_{\mathcal O_H}\Bbb F_4$ with $\Phi$ as structural map.)
Time to get the equation of the curve.
Computers are in this century the weapon of choice, when a lookup in the database goes into the void.
In the above theoretical setup, we use the code:
R.<x> = PolynomialRing(QQ)
K.<s> = NumberField( ( (sqrt(-5) + sqrt(-1))/2 ).minpoly() )
a = (x^2 + 5).roots(ring=K, multiplicities=False)[0]
k = (x^2 + 1).roots(ring=K, multiplicities=False)[0]
j = 320*(1975 + 884*a*k)
E = EllipticCurve_from_j(j)
And E is:
sage: E
Elliptic Curve defined by
y^2 = x^3 + (2142429020160*s^2+818331402240)*x
+ (1803656541954375680*s^2+688935500534251520)
over Number Field in s with defining polynomial x^4 + 3*x^2 + 1
(Code was manually rearranged.)
Now the whole discussion about $\Phi$ is computationally useless, (but structurally needed to have a morphism to "something" in characteristic two, thus giving sense to the story,) since over $\Bbb F_2$ most involved coefficients vanish, giving rise to $y^2=x^3$.
$\square$
The answer is finished, but i tried to find other $j$-values (to search for them), those for which we may obtain after reduction regular elliptic curves over $\Bbb F_4$. The list of these possible CM-discriminants, that
are not discarded from the start (having $a_4$, $a_6$ not zero modulo $(2)$) was obtained via:
sage: KK.<sqrt5> = QuadraticField(5)
sage: RZ.<Y> = PolynomialRing(ZZ)
sage: RF2.<X> = PolynomialRing(GF(2))
sage: for j in cm_j_invariants(K):
....: F = RQ(j.minpoly())
....: f = F(X) # reduction modulo two
....: if f.is_irreducible() and f.degree() == 2:
....: print(f'j = {j} is among {F.roots(ring=KK, multiplicities=False)}')
....: E = EllipticCurve_from_j(j)
....: print(E, '\n')
We get:
j = 85995*s^2 + 33480 is among [85995/2*sqrt5 - 191025/2, -85995/2*sqrt5 - 191025/2]
Elliptic Curve defined by
y^2 = x^3 + (49727382705*s^2+18996249195)*x
+ (-6410928347984160*s^2-2448758247517890)
over Number Field in s with defining polynomial x^4 + 3*x^2 + 1
j = -85995*s^2 - 224505 is among [85995/2*sqrt5 - 191025/2, -85995/2*sqrt5 - 191025/2]
Elliptic Curve defined by
y^2 = x^3 + (-49727382705*s^2-130185898920)*x
+ (6410928347984160*s^2+16784026796434590)
over Number Field in s with defining polynomial x^4 + 3*x^2 + 1
j = 16554983445*s^2 + 43341513480 is among [16554983445/2*sqrt5 + 37018076625/2, -16554983445/2*sqrt5 + 37018076625/2]
Elliptic Curve defined by
y^2 = x^3 + (-1838500851256815240495*s^2-4813257716935987046805)*x
+ (-45371836401785381434095138749760*s^2-118785009831873858847333508631390)
over Number Field in s with defining polynomial x^4 + 3*x^2 + 1
j = -16554983445*s^2 - 6323436855 is among [16554983445/2*sqrt5 + 37018076625/2, -16554983445/2*sqrt5 + 37018076625/2]
Elliptic Curve defined by
y^2 = x^3 + (1838500851256815240495*s^2+702244836834458674680)*x
+ (45371836401785381434095138749760*s^2+17330499373482285454951907617890)
over Number Field in s with defining polynomial x^4 + 3*x^2 + 1
And over $\Bbb F_4=\Bbb F_2[u]$ we get respectively the curves (recalling $\Phi(s^2)=u+1$) $E_1$, $E_2$, $E_3=E_1$, $E_4=E_2$, with:
$$
\begin{aligned}
E_1 &:\qquad y^2 = x^3 + ux\ ,\\
E_2 &:\qquad y^2 = x^3 + (u+1)x \ .
\end{aligned}
$$
They are singular because of $x^3 +ux=x(x+u+1)^2$, and $x^3+(u+1)x=x(x+u)^2$.
Best Answer
For this particular elliptic curve, the Mordell-Weil rank is equal to $0$. This is equivalent to the fact that $1$ is not a congruent number, which can be proved with elementary methods.
Thus we can completely determine: $E(\Bbb Q) = \{(-1, 0), (0, 0), (1, 0), \infty\} = E[2]$.
Thus the set of all points $P$ with $[2]P \in E(\Bbb Q)$ is simply $E[4]$, which has $16$ points and can be totally determined by solving the equation $[4]P = 0$.
We can compute $E[4]$ by solving equations $[2]P = Q$ for each $Q \in E[2]$. For $Q = \infty$ we get just $E[2]$ as four solutions.
The doubling formula for this curve is $x([2]P) = \frac{x^4 + 2x^2 + 1}{4(x^3 - x)}$, where $x = x(P)$.
Thus for example the $x(P)$ values for the four solutions to $[2]P = (-1, 0)$ are the roots of the polynomial $(x^4 + 2x^2 + 1) + 4(x^3 - x) = 0$, which are $-1 \pm \sqrt 2$, each with multiplicity $2$ since there are two $y(P)$ values for each $x(P)$ value.
Calculations for the other two are similar.