, 2021

The only solution is $(a, b, c) = \left(0, \frac{1}{2}, \frac{3}{2}\right)$.
Note that $(2^a - 1)^2 \ge 0$, $(2^b - \sqrt{2})^2 \ge 0$ and $(2^c - \sqrt{8})^2 \ge 0$. These yield
$$2^{2a} + 1 \ge 2^{a+1},$$
$$2^{2b} + 2 \ge 2^{b+\frac{3}{2}},$$
$$2^{2c} + 8 \ge 2^{c+\frac{5}{2}}.$$
Multiplying these inequalities, we obtain
$$(2^{2a} + 1)(2^{2b} + 2)(2^{2c} + 8) \ge 2^{a+b+c+1+\frac{3}{2}+\frac{5}{2}} = 2^{a+b+c+5}.$$
This shows all equalities should hold. This means $(a, b, c) = \left(0, \frac{1}{2}, \frac{3}{2}\right)$.