John Cremona has some explicit code for calculating the 2-torsion points of curves in the general Weierstrass [a1,a2,a3,a4,a6] format, and part of that formula is finding rational roots to the cubic equation
(1) $ P(x,[W]) = 4(x^3+e_{a2}x^2+e_{a4}x+e_{a6})+(e_{a1}x+e_{a3})^2)$
where [a1,a2,a3,a4,a6] are from the Weierstrass form W.
If you stick in a curve of the form [0,a,0,b,0] then (1) becomes
(2) $4x^3 + 4ax^2 + 4bx = 0$ (to find the 2-torsion points)
and it is easy to see that $x=0$ is one torsion point. The other 2 are the quadratic roots of $x^2 + 4ax + b$ which need to be rational, if three 2-torsion points are to be found.
However if we use the Tate form above of $[1-c,-b,-b,0,0]$ and stick it into (1) we derive
(2) $P(x,b,c) = 4x^3 + (c^2 - 2c + (-4b + 1))x^2 + (2bc - 2b)x + b^2$
and we want rational roots of this cubic for x.
Maxima tells us that the 3 roots of this cubic in [b,c] is
$x_1=\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}}\right)\,\left({{b\,
\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1
\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^
2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,
b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{
1728}}\right)^{{{1}\over{3}}}-{{\left({{\sqrt{3}\,i}\over{2}}+{{-1
}\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\,
b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\,
\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+
16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}}
-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right)
}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}}
\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$
$x_2 = \left({{\sqrt{3}\,i}\over{2}}+{{-1}\over{2}}\right)\,\left({{b\,
\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1
\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^
2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,
b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{
1728}}\right)^{{{1}\over{3}}}-{{\left({{-1}\over{2}}-{{\sqrt{3}\,i
}\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\,
b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\,
\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+
16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}}
-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right)
}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}}
\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$
and
$ x_3 = \left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+
\left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}
}}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(
\left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c
-1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}-{{{{\left(c-1\right)
\,b}\over{6}}-{{\left(4\,b-c^2+2\,c-1\right)^2}\over{144}}}\over{
\left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+
\left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}
}}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(
\left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c
-1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1
}\over{12}}$
I really don't see a simple form here for rational roots. unless b and c take very specific values so that one of $x_1$, $x_2$ or $x_3$ become rational.
I did also consider the elliptic curve invariants $c_4$ and $c_6$ equivalency, but it means moving from a sextic to a quartic equation for the two variables (b,c --> a,b)
(continuing) after reading up a bit more on 2-isogenies and Magma defining fields, the following Magma code and results is your answer, no, there is no simple elliptic curve:
F<a,b,c>:=FunctionField(Rationals(),3);
E:=EllipticCurve([1-c,-b,-b,0,0]);
E;
Elliptic Curve defined by y^2 + (-c + 1)xy - by = x^3 - bx^2 over Multivariate rational function field of rank 3 over Rational Field
E1, f := IsogenyFromKernel(E, DivisionPolynomial(E, 2));
E1;
Elliptic Curve defined by y^2 + (-c + 1)xy - by = x^3 - bx^2 + (-5b^2 + 5/2bc^2 + 5/2bc - 5b - 5/16c^4 + 5/4c^3 - 15/8c^2 + 5/4c - 5/16)x + (-3b^3 + 9/4b^2c^2 + 7/2b^2c + 10b^2 - 9/16bc^4 + 1/4bc^3 + 21/8bc^2 - 15/4bc + 23/16b + 3/64c^6 - 9/32c^5 + 45/64c^4 - 15/16c^3 + 45/64c^2 - 9/32c + 3/64) over Multivariate rational function field of rank 3 over Rational Field
f;
Elliptic curve isogeny from: CrvEll: E to CrvEll: E1
taking (x : y : 1) to ((x^4 + (-b + 1/4c^2 - 1/2c + 1/4)x^3 + (b^2 - 1/2bc^2 + 1/2b + 1/16c^4 - 1/4c^3 + 3/8c^2 - 1/4c + 1/16)x^2 + (-1/2b^2c - 3/2b^2 + 1/8bc^3 - 3/8bc^2 + 3/8bc - 1/8b)x + (3/4b^3 + 1/16b^2c^2 - 1/8b^2c + 1/16b^2)) / (x^3 + (-b + 1/4c^2 - 1/2c + 1/4)x^2 + (1/2bc - 1/2b)x + 1/4b^2) : (x^6y + (-2b + 1/2c^2 - c + 1/2)x^5y + (b^2c - b^2 - 1/2bc^3 + 3/2bc - b + 1/16c^5 - 5/16c^4 + 5/8c^3 - 5/8c^2 + 5/16c - 1/16)x^5 + (5/2bc - 5/2b)x^4y + (-1/2b^3c + b^3 + 3/8b^2c^3 - 11/8b^2c^2 - 5/2b^2c + 7/2b^2 - 3/32bc^5 + 1/2bc^4 - 17/16bc^3 + 9/8bc^2 - 19/32bc + 1/8b + 1/128c^7 - 7/128c^6 + 21/128c^5 - 35/128c^4 + 35/128c^3 - 21/128c^2 + 7/128c - 1/128)x^4 + 5b^2x^3y + (1/2b^3c^2 + 9/4b^3c - 5b^3 - 1/4b^2c^4 + 11/16b^2c^3 - 9/16b^2c^2 + 1/16b^2c + 1/16b^2 + 1/32bc^6 - 3/16bc^5 + 15/32bc^4 - 5/8bc^3 + 15/32bc^2 - 3/16bc + 1/32b)x^3 - 5b^3x^2y + (-5/4b^4c + 7/2b^4 + 1/8b^3c^3 - 9/16b^3c^2 + 3/4b^3c - 5/16b^3 + 3/64b^2c^5 - 15/64b^2c^4 + 15/32b^2c^3 - 15/32b^2c^2 + 15/64b^2c - 3/64b^2)x^2 + (2b^4 - 1/2b^3c^2 + 1/2b^3c)xy + (-b^5 + 5/8b^4c^2 - 7/8b^4c + 1/4b^4 + 1/32b^3c^4 - 1/8b^3c^3 + 3/16b^3c^2 - 1/8b^3c + 1/32b^3)x - 1/2b^4cy + (11/32b^5c - 1/16b^5 + 1/128b^4c^3 - 3/128b^4c^2 + 3/128b^4c - 1/128b^4)) / (x^6 + (-2b + 1/2c^2 - c + 1/2)x^5 + (b^2 - 1/2bc^2 + 2bc - 3/2b + 1/16c^4 - 1/4c^3 + 3/8c^2 - 1/4c + 1/16)x^4 + (-b^2c + 3/2b^2 + 1/4bc^3 - 3/4bc^2 + 3/4bc - 1/4b)x^3 + (-1/2b^3 + 3/8b^2c^2 - 3/4b^2c + 3/8b^2)x^2 + (1/4b^3c - 1/4*b^3)x + 1/16b^4) : 1)
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Best Answer
Is this system of equations satisfactory for Question 1:
$x+y+x \times z+y \times z=6400$,
$x \times y \times z=6561$?
It seems that all curves with torsion groups containing $\mathbb{Z}/4\mathbb{Z}$ can be obtained in this way. By taking $x+y+x \times z+y \times z= d$, $x \times y \times z=-cd$, we get the elliptic curve $[1,-c/d,-c/d,0,0]$ with a point $[0,0]$ of order $4$.