If $F(z)=z^\gamma$ is an algebraic function (over $\mathbb Q$), then $F'(z)$ is also an algebraic function (over $\mathbb Q$), and thus $F'(1)$ is an algebraic number (over $\mathbb Q$). Therefore, $z^\gamma$ algebraic implies $\gamma$ algebraic. Equivalently, if $\gamma$ is not algebraic, then $z^\gamma$ is not algebraic.
In some cases, one knows a Puiseux series expansions of a function $F(z)$; such an expansion is unique and is of the type $F(z) = \sum_{k\geq k_0} f_k \times \big((z-z_0)^{r} \big)^k \,.$ What is more, the theory of Newton polygon for algebraic functions implies that the exponent $r$ in this Puiseux expansion is then a rational number. This gives a nice way to test if $F(z)$ is algebraic or not!
Therefore, one can say that $z^\gamma$ is an algebraic function if and only if $\gamma$ is a rational number (the same reasoning proves that $z^{\sqrt 2}$ is not an algebraic function). [It just requires a little lemma showing that $z^\gamma$ cannot be rewritten as an infinite sum of $z^{rk}$, which follows easily by considering the limit when $z\rightarrow 0$.]