Solution:
Let $y = x^{2} - 4x + 1$.

The expression $x^{2} - 4x + 1$ is a quadratic in $x$.

The minimum value of $y$ occurs at $x = -\frac{b}{2a} = \frac{4}{2} = 2$.

At $x = 2$:
$$y = (2)^{2} - 4 \cdot 2 + 1 = 4 - 8 + 1 = -3$$

As $x \to \pm \infty$, $y \to +\infty$.

Therefore, $y \in [-3, \infty)$.

So, $f(x) = 2^{y} = 2^{x^{2} - 4x + 1}$ takes all values $2^{t}$ where $t \in [-3, \infty)$.

Thus, the range is $[2^{-3}, \infty) = [\frac{1}{8}, \infty)$.