Triplets satisfying $(a^3+b)(b^3+a)=2^c$ 
Find the number of triplets $(a,b,c)$ satisfying $(a^3+b)(b^3+a)=2^c$, where $a,b,c\in \mathbb{N}$  

A trivial solution is $(1,1,2)$. I think there aren't any other such triplets, so I've been trying to prove that but haven't been able to do so.
How can other solutions be found?
 A: shaurya gupta already noted the solution $(a,b,c) = (1,1,2)$;
rogerl found another pair, $(3,5,12)$ and $(5,3,12)$, and conjectured
that there are no others.  I prove that this conjecture is correct.
[The question required that $a,b,c \in \mathbb N$; this notation
"$\mathbb N$" is sometimes used for nonnegative integers,
but I assume that zero is not allowed here, else
$(a,b,c) = (0,2^r,4r)$ is a solution for every nonnegative integer.]
rogerl shows that $a,b$ must be odd.
Assume without loss of generality that $a \leq b$, and write
$$
a^3 + b = 2^d, \quad
b^3 + a = 2^e 
$$
for some integers $d,e$ with $d \leq e$ and $c = d+e$.
Then $b = \lfloor 2^{e/3} \rfloor$, which  makes it easy to check that 
the three known solutions are the only ones with $e \leq 15$ 
(for each $e$, solve for $b$, recover $a = 2^e - b^3$,
and check whether $0 < a \leq b$).  Thus any other solution
must have $e>15$, and therefore $d>5$ (because $b^3+a < (a^3+b)^3$).
Now $a,b$ satisfy $a^3 \equiv -b$ and $b^3 \equiv -a$ mod $2^d$.
Therefore $a^9 \equiv a \bmod 2^d$.  Since $a$ is odd it follows that $a^8 \equiv 1$,
so $a \equiv \pm 1 \bmod 2^{d-3}$.  Hence $a=1$, else $a \geq 2^{d-3}-1$
and $a^3 > 2^d$.  But then $b=2^d-1$, and (since $d>5$) we have reached a
contradiction because $b^3+a$ is strictly between $2^{3d-1}$ and $2^{3d}$.
QED
P.S. For the record:
@ $a,b$ are odd because if a power of $2$ is the sum of two positive integers
then their 2-valuations are equal, and this would not be possible with
$a,b$ even.
@ One proof of the implication
$a^8 \equiv 1 \bmod 2^d \Longrightarrow a \equiv \pm1 \bmod 2^{d-3}$ is to
write $\pm a = 1 + 4n$ and expand $(a^8-1)/32$ as a polynomial in $n$.
A: Note that $a$ and $b$ must be odd and relatively prime. For they must have the same parity, and if they are both even, say $a = 2^rt$ and $b = 2^su$ for $t$ and $u$ odd, then (assuming wlog that $1\le r\le s$)
$$(a^3+b)(b^3+a) = (2^{3r}t + 2^su)(2^{3s}u+2^rt) 
    = 2^r(2^{3r}t + 2^su)(2^{3s-r}u+t) = 2^c.$$
Thus $(2^{3r}t + 2^su)(2^{3s-r}u+t)$ is a power of $2$, so that $3s-r=0$ and $r=3s$. But then the first factor cannot be a power of $2$.
The only solution with $a, b\le 10,\!000$ is $\{3,5,12\}$ (and of course $\{5,3,12\}$).
Given the relative paucity of cubes, I'd be surprised to find additional solutions.
