Here's a way to construct a solution for each triple $(m,n,t)$ of pairwise coprime integers.

Start by choosing your favorite $a,b,c$ so that $a+b=c$ and write down the prime factorizations $$a=\prod_{i=1}^k p_i^{\alpha_i} \ , \ b=\prod_{i=1}^k p_i^{\beta_i} \ , \ c=\prod_{i=1}^k p_i^{\gamma_i}.$$ Now, by the Chinese remainder theorem find large enough $\delta_i$'s so that $$\alpha_i-\delta_i\equiv 0\pmod{m}$$
$$\beta_i-\delta_i\equiv 0\pmod{n}$$
$$\gamma_i-\delta_i\equiv 0\pmod{t}$$ and denote $D=\prod_{i=1}^k p_i^{\delta_i}$. You have $$\frac{a}{D}+\frac{b}{D}=\frac{c}{D}$$ and so $x^{-m}+y^{-n}=z^{-t}$ for some integers $x,y,z$.