A diophantine equation with only "titanic" solutions

The equation $313(x^3+y^3) = t^3$ is equivalent to finding,

$$x^3+y^3=313^2z^3\tag1$$

As Achille Hui points out, Noam Elkies found the solution with $83$-digits,

$$\small x_0 = 355507307842882624593086325021133856149447336710120844428552934573043094018915 289363\\ \small y_0 = -354602746692986709129018423204648314355484458881941451025238387384142099383045 862152 \\ \small z_0 =1517122651849438712721950935044230084378368307868200665761294465082177989014675811$$

Update: Curiously, $3(x_0+y_0) = (3\cdot 7\cdot 8273\cdot 64072783\cdot 125303678787043)^3$.

However, the OP wants them positive. Using the method also discussed in this post, given an initial solution to,

$$ax^3+by^3 = cz^3\tag2$$

then a new one can be derived as,

$$a(-bxy^3-cxz^3)^3 + b(ax^3y+cyz^3)^3 = c(-ax^3z+by^3z)^3\tag3$$

We can use $(3)$ iteratively to find an infinite number of solutions. We have,

$$x_0,y_0,z_0 = +,\,-,\,+\\ x_1,y_1,z_1 = +,\,-,\,-\\ x_2,y_2,z_2 = +,\,-,\,+\\ x_3,y_3,z_3 = +,\,-,\,-\\ x_4,y_4,z_4 = \color{red}{+,\,+,\,+} $$

and so on. So the fourth iteration is all positive. Approximately,

$$x_4 = 1.908757×10^{21389}\\ y_4 = 4.955536×10^{21389}\\ z_4 = 1.095063×10^{21388}$$

They are too long to explicitly write down, but if you have Mathematica, you can retrace the steps taken and see those numbers in all their glory.

The curve $a^3+b^3=N$ is birationally equivalent to the elliptic curve $y^2=x^3-432N^2$ with \begin{equation} a=\frac{36N+y}{6x} \hspace{2cm} b=\frac{36N-y}{6x} \end{equation}

For $N=313^2$, the curve has rank $1$ and generator $G=(x_0,y_0)$ with

$x_0$=426235512202934545020503360093256801131707221692968586587468/216170759226021502298882345008844433022529079715666681

$y_0$=278275087731298331021683520315726613848790652329435004093249928083293904849586928211092140/100506794432879496007544646276171310440319758686599267034949687655666070652158579

which give the solution Noam Elkies tabulated.

The curve has no torsion points so all rational solutions come from points of the form $mP, \, \, m=1,2,\ldots$.

Looking at the above transformations, a positive solution will only arise when $|y| < 36N$ since $x>0$ always.

Experiments show that this first happens when $m=9$ giving a solution with roughly 6770 digits