8. The equation $\frac{1}{x^{2}}+\frac{1}{y}+\frac{1}{z}=\frac{4}{3}$ has $\qquad$ integer solutions.
Consider the integer solutions of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{4}{3}$, assuming $a \geqslant b \geqslant c$. Clearly, there can be at most $c0$, then $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c}$. From $\frac{4}{3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{3}{c} \Rightarrow c \leqslant \frac{9}{4} \Rightarrow c=1,2$. When $c=1$, $\frac{1}{3}=\frac{1}{a}+\frac{1}{b} \leqslant \frac{2}{b} \Rightarrow 3\frac{1}{3} \Rightarrow a=1,2$. When $a=2$, $c=-6$. In this case, the equation has 1 solution: $(2,1,-6)$.
Therefore, the solutions that satisfy the original equation are $( \pm 1,4,12),( \pm 1,12,4),( \pm 2,1,12),( \pm 2,12,1),( \pm 1,6,6),( \pm 1,2,-6)$, $( \pm 1,-6,2)$, a total of 14 sets.