What is the difference between Shannon entropy and saying that tossing a 6-sided die 100 times has more than 256-bit entropy? I'm confused.

I thought that tossing a 6-sided die 100 times had a greater than 256-bit entropy because $6^{99} &lt; 2^{256} &lt; 6^{100}$. (A similar concept appeared in this XKCD comic, where choosing four random words from a dictionary of presumably 2048 words has a 44-bit entropy, presumably because $2048^4 = 2^{44}$.)

On the other hand, the Shannon entropy of a 6-sided die tossed 100 times is $-6 × 1/6 × \log_2(1/6) = 2.5849625007$ bits. (According to the comments by Sakamaki Izayoi to my other question.)

Are these two different concepts of "entropy" entirely? If so, what's the difference and if not, what am I missing?

I read this thread but I'm still confused.
On the other hand, the Shannon entropy of a 6-sided die tossed 100 times is $-6 × 1/6 × \log_2(1/6) = 2.5849625007$ bits.


That is wrong: $-6\cdot\frac16\cdot\log_2\frac16$ is the entropy of a single die roll. Assuming the $100$ die rolls are independent, you can simply sum the entropies of the individual rolls to obtain
$$ 100 \cdot\left(-6 \cdot\frac16\cdot\log_2\frac16\right) \approx 258.49625 \text, $$
which is precisely $\log_2(6^{100})$ as expected.