Solution:
Setting $y = 2^{x}$, we have to find all values of $a$ such that the equation
$$y^{2} - (a^{2} + 3a - 2) y + 3a^{3} - 2a^{2} = 0 \Longleftrightarrow (y - a^{2})(y - 3a + 2) = 0$$
has exactly one positive root.

Obviously $a = 0$ is not a solution. For $a \neq 0$ the equation has a positive root $y_{1} = a^{2}$. It is unique if either $y_{2} = 3a - 2 \leq 0$ or $y_{2} = y_{1}$.

In the first case we obtain $a \leq \frac{2}{3}$ and in the second one we have $a^{2} = 3a - 2$, whence $a_{1} = 1$ and $a_{2} = 2$.

Finally, $a \in (-\infty, 0) \cup (0, \frac{2}{3}) \cup \{1\} \cup \{2\}$.