Suppose that \(s_m\) is a power of 2 for some non-negative integer \(m\). In particular, \(2s_m = 2^{m+2} - m^2 - m - 2\) is a power of two. On the other hand, observe that if \(m \geq 4\), then \(2^m > m^2\), \(2^{m-1} > m\), and \(2^{m-2} > 2\). It follows that $$2^{m+1} > 2^m + 2^{m-1} + 2^{m-2} > m^2 + m + 2$$Thus, $$2^{m+1} = 2^{m+2} - 2^{m+1} < 2^{m+2} - m^2 - m - 2 = 2s_m < 2^{m+2}$$Therefore, if \(m \geq 4\), then \(2s_m\) is not a power of 2. Finally, by checking for \(m = 0, 1, 2, 3\), we obtain that the only solutions are \(m = 0, 1, 2\), and we are done.