Saudi Arabian Mathematical Competitions

We have
$$s_{m} = \sum_{k=0}^{m} \left(2^{k} - k\right) = 2^{m+1} - 1 - \frac{m(m+1)}{2}$$
We prove that for $m \geq 3$, we have $2^{m} < s_{m} < 2^{m+1}$. This inequality is equivalent to
$$2^{m} < 2^{m+1} - 1 - \frac{m(m+1)}{2} < 2^{m+1}$$
The right inequality is obvious. The left inequality is equivalent to
$$1 + \frac{m(m+1)}{2} < 2^{m}$$
that is
$$2 + 2 \frac{m(m+1)}{2} < 2^{m+1}$$
or
$$2\left(\binom{m+1}{0} + \binom{m+1}{2}\right) < 2^{m+1}$$
and we are done.

For $m=0$, we have $s_{0} = 1 = 2^{0}$.
For $m=1$, we have $s_{1} = 2 = 2^{1}$.
For $m=2$, we have $s_{2} = 4 = 2^{2}$.
The solutions are $m \in \{0, 1, 2\}$.