If such a set $A$ of size $m$ exists, all its admissible pairwise sums must lie in its complement, thus
$$
\binom{m}{2} \leq 2^n-m,
$$
which gives $$m\leq 2^{(n+1)/2}\qquad (1)$$.


**Edit:** *The upper bound is wrong. See the wonderful comments and examples below the question and in the answers. I can only accept one answer, unfortunately.*


Since $a+b=c$ is the same as $a+b+c=0$ in characteristic 2, a randomly chosen set of size roughly $2^{n/3}$ will have a solution with some constant positive probability.


For simplicity let $n=2k$. Consider the nonzero elements in $\mathbb{F}\_{2^k}$ as representing the vectors in $\mathbb{F}\_2^k$
and take "vectors" of the form $(x,1/x)$ in $\mathbb{F}\_2^n.$
Take $$(x,1/x)+(y,1/y)=(z,1/z)$$
equivalently $$x+y+z=0,$$ and $$1/x+1/y+1/z=0$$
as the simultaneous equations we must avoid.


By the Newton-Girard formulas we equivalently want to avoid
$$x+y+z=0,$$ and $$x^2+y^2+z^2=0$$
but in characteristic 2 the second equation is just the equare of the first.


This seems to say, if I pick a primitive element $\alpha$ in the subfield defined above and define
$v(i)=(\alpha^i,\alpha^{-i}+\beta)$ where $\beta$ is a nonzero element in $\mathbb{F}\_{2^n} \setminus\mathbb{F}\_{2^k}$, the vectors 
$$\{ v(i): 0\leq 2^k-2\}$$ form the type of set we want with $m=2^{n/2}-1.$


**Question:** Is there a better construction which approaches (1).



Largest $A\subset \mathbb{F}_2^n$ such that no two $a\neq b$ in $A$ add to an element of $A.$
The upper bound is wrong. The problem is that there is no need for the pairwise sums to be distinct. However it is true that there must be at least $m-1$ distinct sums, which gives $m \leq 2^{n-1}$.


This is sharp as we can take $A$ to consist of all elements with nonzero first entry, so all the sums of two elements of $A$ have zero first entry and so do not lie in $A$.


