Weierstrass form of genus one $y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0$ Related to the n-conjecture.
We are looking for Weierstrass form and map from it of the genus one curve:
$$ y^{10} z^{30} - 8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10} - 64=0 $$
It is of degree $40$ and has only five monomials.

Added We suspect there are finite number of rational points (likely none).
So the Weierstrass form may be over number field with monic defining polynomial with integer coefficients. The rank must be positive in addition. This relaxation might help Magma (we don't have it).

Maple gave weird error.
In machine readable form:
y^10*z^30-8000*y^4*z^20+1600*z^10*y^2+12800000*z^20-64

 A: (a) The given curve is irreducible over $\mathbb Q$, but reducible over $\mathbb Q(\sqrt[5]{2})$. So it does not have a Weierstrass form, and each rational point has to be a singularity. The curve however does not have rational singularities (there are two on the projective completion though). 
(b) From the latest comments by the OP it has become clear that he wants a Weierstrass model for a component of the curve. Set $\alpha=\sqrt[5]{2}$. Then the curve given by $y^2z^6 + 20\alpha^2z^4 -2\alpha=0$ is one of the algebraically conjugate components. Over $\mathbb Q(\alpha,\sqrt[4]{20})$ this is easily brought into Weierstrassform by hand.
What follows was written before I realized that the curve isn't absolutely irreducible. So it may help in such situations, though it has little value (and lacking correctness) here.
The following works: Set $F(y,z)=y^{5} z^{3} - 8000 y^{2} z^{2} + 12800000 z^{2} + 1600 y z - 64$. Then your elliptic curve $C$ is the solution set of $F(y^2,z^{10})=0$. Let $C'$ and $X$ be the solution sets of $F(y^2,z)=0$ and $F(y,z)=0$. So there are natural covers $C\to C'\to X$. Note that $C'$ has genus $1$, and $X$ has genus $0$.
Maple easily finds a rational parametrizaton of $X$. After some simplifications we see that $y=s^3-20s$, $z=\frac{4}{s^5}$ is a parametrization of $F(y,z)=0$.
So a Weierstrass form of $C'$ is given by $E':y^2=s^3-20s$, and $\alpha':E'\to C'$, $(s,y)\mapsto(\frac{4}{s^5},y)$ is birational. Let $\alpha:E\to C$ be a birational map (defined over some number field) of a Weierstrass model $E$ of $C$, and $\psi:C\to C'$ be the natural map $(z,y)\mapsto (z^{10},y)$.
Then $\alpha'^{-1}\circ\psi\circ\alpha$ is an isogeny $E\to E'$ of degree $10$.
If a non-singular model of $C$ had a rational point, then $E$ and $\alpha$ could be defined over $\mathbb Q$, and $E'$ would have an isogeny of degree $10$ defined over $\mathbb Q$. However, the $5$-division polynomial of $E'$ does not have a quadratic factor over $\mathbb Q$, so $E'$ and then also $E$ has no rational isogeny of degree $5$, and therefore also none of degree $10$.
In particular, as expected by the OP, $C$ has only finitely many rational points.
How to explicitly find $E$ and $\alpha$? Using (for instance) Sage one can find $E$ and an isogeny $\mu:E\to E'$ of degree $10$. (First find the degree $10$ isogeny $E'\to E$, then take its dual.) Finding $\alpha$ amounts to expressing a generic point $(z,y)\in C$ in terms of the generic point $(u,v)\in E$. This should be doable via $(z^{10},y)=\psi((z,y))=\alpha'(\mu((u,v))$, because we can compute everything on the right hand side. 
