Is there a way to express $(2^k + 1)^{k+1}$ expanded...
The first few values for $k+1=2,3,4,5,6,\ldots$ are:
$ k+1=2: 1+2^{2 k}+2^{1+k} $ $ k+1=3: 1+3\ 2^k+3\ 2^{2 k}+2^{3 k} $ $ k+1=4: 1+2^{4 k}+2^{2+k}+3\ 2^{1+2 k}+2^{2+3 k} $ $ k+1=5: 1+5\ 2^k+5\ 2^{4 k}+2^{5 k}+5\ 2^{1+2 k}+5\ 2^{1+3 k} $ $ k+1=6: 1+15\ 2^{2 k}+15\ 2^{4 k}+2^{6 k}+3\ 2^{1+k}+5\ 2^{2+3 k}+3\ 2^{1+5 k} $ $ k+1=7: 1+7\ 2^k+21\ 2^{2 k}+35\ 2^{3 k}+35\ 2^{4 k}+21\ 2^{5 k}+7\ 2^{6 k}+2^{7 k} $
By the binomial theorem, $$(2^k+1)^{k+1}=\sum_{i=0}^{k+1}\binom{k+1}{i}2^{ik}$$
Take Pascal's triangle and multiply each term in a row by increasing powers of $2^k$.
If you really want to group with the powers of 2 contained in the coefficients, you can use the divisibility criterion of Kummer. http://en.wikipedia.org/wiki/Binomial_coefficient#Divisibility_properties
Let $C(m, n)$ be the number of carries when performing the binary addition of $m$ and $n$. You can rewrite as $$\sum_{i=0}^{k+1}\frac{\binom{k+1}{i}}{2^{C(i, k+1-i)}}2^{ik+C(i, k+1-i)},$$ such that the coefficients are odd.
For example, for $k+1=4$:
$000_b+100_b\rightarrow$ no carry, $001_b+011_b\rightarrow 2$ carries, $010_b+010_b\rightarrow 1$ carry...
$$\begin{align}1.2^0+ 4.2^3 + 6.2^6+ 4.2^9+ 1.2^{12}&=\frac1{2^0}.2^{0+0}+\frac4{2^2}.2^{3+2}+\frac6{2^1}.2^{6+1}+\frac4{2^2}.2^{9+2}+\frac1{2^0}.2^{12+0}\\&=1+2^5+3.2^7+2^{11}+2^{12}\end{align}$$