Scary powers of 2

Let $k = 2^a b$ where $a \ge 0$ and $b$ is odd. I claim that $2^{2^k-1} - 2^{k-1} - 1$ is divisible by $N_a = 2^{2^a}+1 = 2^{k/b} + 1.$

Let's look mod $N_a.$ First, note that $2^{k-1} \equiv \left( 2^{k/b} \right)^b 2^{-1} \equiv (-1)^b 2^{-1} \equiv -2^{-1} \equiv -1/2.$ Then look at the first term: $2^{2^k-1} \equiv \left( 2^{2^a} \right)^{2^{k-a}} 2^{-1} \equiv (-1)^{2^{k-a}} 2^{-1} \equiv 1/2.$ So the original number is congruent to $1/2 -(-1/2) - 1 \equiv 0$ mod $N_a.$

I'll leave to the reader the proof that $N_a$ is strictly smaller than $2^{2^k-1}-2^{k-1}-1$ for $k > 2.$ This shows that $2^{2^k-1}-2^{k-1}-1$ is composite.

${ 2 ^ { 2 } } ^ { 3 } -1 - 2 ^ { 2 } - 1 = 128 - 4 - 1 = 123 = 3 \times 61$ .

So it is always composite , so the answer is $\boxed { \text { Yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! } }$