I think that this is equivalent to a known open question. Here are the details. For $K:=\mathbb{F}$$_{2^n}$, the function $f:y\mapsto y+y^2:K\to K$ is $\mathbb{F}$$_2$-linear, and its kernel $\{0,1\}$ has dimension 1. The image is therefore of dimension $n-1$, and for $z$ in the image, the fiber $f^{-1}(z)$ has exactly 2 elements.
Hence, to prove that $y^2+y=x^k+ax$ has exactly $2^n$ solutions for some fixed $a\in K$, we have to show that $|\{x\in K|x^k+ax\in \mathrm{Im}(f)\}|=2^{n-1}$.
Since $\sigma:y\mapsto y^2$ is a generator of the Galois group of $K/\mathbb{F}$$_2$, Hilbert's Theorem 90 (in additive form) says that $z\in \mathrm{Im}(f)$ if and only if $\mathrm{Tr}(z)=0$, where $\mathrm{Tr}$ stands for the trace map from $K$ to $\mathbb{F}$$_2$.
So the problem is equivalent to showing that there exists an $a\neq 0$ in $K$ such that $|\{x\in K|\mathrm{Tr}(x^k+ax)=0\}|=2^{n-1}$. In other words, we would like to show that there exists a nonzero $a\in K$ such that $$ S_k(a):=\sum_{x\in K}(-1)^{\mathrm{Tr}(x^k+ax)}=0. $$
Apparently, this question was addressed in the coding community. In detail, in [1, p. 258], the following conjecture (of Helleseth) is mentioned:
``Conjecture 3. For any $m$ and $k$ such that $\mathrm{gcd}(2^m-1,k)=1$, the sum $\sum_{x\in\mathbb{F}_{2^m}}(-1)^{\mathrm{Tr}(x^k+ax)}$ is null for at least one nonzero $a$.'' (Note that $n$ in the current question is $m$ in [1]).
It seems that in [1, Corollary 1, p. 253], Conjecture 3 is proved for even $m$ and for certain values of $k$ (the ``Niho exponents,'' defined on p. 252 of [1]).
Interestingly, at least at a first glance it seems that [1] has nothing to say on $k\in\{1,\ldots,2^{n-1}\}$, but to me it seems that this case is trivial (am I missing something?): Consider a normal basis for $K/\mathbb{F}$$_2$, that is, a basis $B$ consisting of an orbit of an element $\gamma\in K$ under the Galois group of $K/\mathbb{F}_2$ (the $i$th element of $B$ is $b_i:=\gamma^{2^i}$ for $i\in\{0,\ldots,n-1\}$).
From the linearity of the trace and the fact that the trace is onto, we must have $\mathrm{Tr}(b)=1$ for at least one element $b\in B$, and from $\mathrm{Tr}(b^2)=\mathrm{Tr}(b)$ we then have $\mathrm{Tr}(b)=1$ for all $b\in B$. So the trace of an element in $K$ is just the modulo-2 sum of the coefficients in its decomposition according to the basis $B$.
Let $a$ be any element in the trace-dual basis of $B$, say $\mathrm{Tr}(ab_i)=\delta_{i,0}$. Then for $k=2^j$, if we write $x=\sum_i \alpha_i b_i$, we get: $\mathrm{Tr}(ax)=\alpha_0$, $\mathrm{Tr}(x^k)=\mathrm{Tr}(x)=\sum \alpha_i$ (sum in $\mathbb{F}_2$). These agree for half of the $x\in K$, as required.
That's about it. I hope at least some of this makes sense :) I also hope that the original person asking this question didn't actually want to solve the above open question by converting it to a question about curves, for then this is answer is useless.
[1] P. Charpin, ``Cyclic codes with few weights and Niho exponents,'' Journal of Combinatorial Theory, Series A 108 (2004) 247--259. (http://www-rocq.inria.fr/secret/Pascale.Charpin/Charpin-jct04.pdf)