Solution:
As $20 \cdot 13 = 2^{2} \cdot 5 \cdot 13$ and $2013 = 3 \cdot 11 \cdot 61$ are relatively prime, $x$, $y$ and $z$ must be nonnegative.

Considering the equation modulo $3$, we observe that $x$ must be odd. Now considering the equation modulo $7$, we obtain $(-1) + (-1)^{y} \equiv 4^{z} \pmod{7}$, which is impossible as the right hand side can only be $1$, $2$ and $4$ modulo $7$.

There are no solutions.