Here's a necessary condition. Write the symmetric polynomial $(x+y)^k - x^k - y^k$ as a polynomial in the elementary symmetric polynomials $x+y$ and $xy$, say $$ (x+y)^k - x^k - y^k = F_k(x+y,xy). $$ Then a necessary condition for $A\in\operatorname{SL}_2(\mathbb Z)$ to be a $k$th power in $\operatorname{SL}_2(\mathbb Z)$ is that the following two conditions hold:

1. $\det A$ is the $k$'th power of an integer, say $\det A=D^k$.
2. The polynomial $$T^k - F_k(T,D) + \operatorname{Trace}(A) $$ has a root in $\mathbb Z$.

The proof is easy enough, since if $B^k=A$, then $\operatorname{Trace}(B)$ is an integer root of the polynomial.

Here's a necessary condition. Write the symmetric polynomial $(x+y)^k - x^k - y^k$ as a polynomial in the elementary symmetric polynomials $x+y$ and $xy$, say $$ (x+y)^k - x^k - y^k = F_k(x+y,xy). $$ Then a necessary condition for $A\in\operatorname{SL}_2(\mathbb Z)$ to be a $k$th power in $\operatorname{SL}_2(\mathbb Z)$ is that the following two conditions hold:

1. $\det A$ is the $k$'th power of an integer, say $\det A=D^k$.
2. The polynomial $$T^k - F_k(T,D) - \operatorname{Trace}(A) $$ has a root in $\mathbb Z$.

The proof is easy enough, since if $B^k=A$, then $\operatorname{Trace}(B)$ is an integer root of the polynomial.

In particular, for the case $k=8$, we have $$ F_k(u,v) = 8u^6v-20u^4v^2+16u^2v^3-2v^4. $$ So every matrix satisfying $B^8=A$ has the property that $\operatorname{Trace}(B)$ is a root of the following polynomial, where $D^k=\det(A)$, $$ T^8 - 8DT^6 + 20D^2T^4 -16 D^3T^2+2D^4 - \operatorname{Trace}(A). $$ In particular, if $B$ is required to have integer entries, then this polynomal has an integer root.