TITLE: Conjecture: Only one Fibonacci number is the sum of two cubes
QUESTION [35 upvotes]: As the title says, I need help proving or disproving that there is only one Fibonacci number that's the sum of two (positive) cubes, $2$. I did a small brute force test with Fibonacci numbers below $10^{15}$ but I couldn't find anything. 
Edit: A possibly fast way to factor sums of cubes is described in another question here. 
I tested Hagen von Eitzen's $F_{3n} - F^3_{n+1}$ up to $n = 200000$ and brute force up to $10^{21}$ but no results. As the numbers grow bigger the chance of a cube gets smaller.

REPLY [6 votes]: There are no further solutions for $n \leq 300$.  Even dropping the
positivity condition the only other solutions are
$F_6 = 8 = 2^3 + 0^3 = 0^3 + 2^3$.
This corroborates Hurkyl's heuristic that it is almost certain
that there are no other Fibonacci numbers that are sums of two cubes.
I doubt that this can be proved, though the search technique described below
can surely be used well past $n=300$, because the main difficulty is
factoring $F_n$, and $F_{300}$ has only about $60$ digits.
To try to express an integer $f$ as $a^3+b^3$, use the factorization
$a^3 + b^3 = (a+b)(a^2-ab+b^2)$.  For each pair $(x,y)$ such that $f=(x,y)$,
check whether $3(4x-y^2)$ is a square, since that's the criterion for
$(x, y) = (a^2-ab+b^2, a+b)$ [the point is that  $4x-y^2 = 3(a-b)^2$].
I did this for $f=F_n$ and all $n \leq 300$ using
this
table of Fibonacci factorizations
(discarding the 100 values of $n$ for which $F_n\bmod 9 \in \{3,4,5,6\}$,
since such a number can never be the sum of two cubes).
This took only ~30 seconds in gp, even though
some candidates $F_n$ in this range have more than a million factors.
Only the known solutions for $F_3=2$ and $F_6=8$ were found.
(also reported at MSE
question 822435.)