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 answer is useless.

1 P. Charpin, ``Cyclic codes with few weights and Niho exponents,''
Journal of Combinatorial Theory, Series A 108 (2004)
247--259.