Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$. 
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.

I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?

If $a$ and $b$ are both even, say $a=2^mA$ and $b=2^nB$ with $A$ and $B$ odd, then without loss of generality $n\geq m$ and so
$$2^c=a^3+b^3=2^{3m}\big(A^3+2^{3(n-m)}B^3\big),$$
from which it follows that
$$2^{c-3m}=A^3+2^{3(n-m)}B^3.$$
Because $A$ and $B$ are positive we see that $2^{c-3m}>1$, and so the left hand side is even. Because $A$ is odd it follows that also $2^{3(n-m)}B^3$ is odd, and so $n=m$. Then for $C=c-3m$ we have
$$A^3+B^3=2^C,$$
thus yielding a new solution with $A$ and $B$ both odd.
If $a$ and $b$ are both odd then you already observed that the factorization
$$2^c=a^3+b^3=(a+b)(a^2-ab+b^2),$$
implies that $a^2-ab+b^2=1$. Viewing this as a quadratic in $a$ shows that
$$a=\frac12\big(b\pm\sqrt{b^2-4(b^2-1)}\big)=\frac12\big(b\pm\sqrt{4-3b^2}\big).$$
For this to be an integer we must have $4-3b^2>0$, so $b=1$ as $b$ is positive by assumption. Then also $a=1$ as $a$ is positive by assumption. This shows that $a=b=1$. It follows that every solution is of the form $a=b=2^m$ for some nonegative integer $m$.
