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?

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 $n$.
