The assumption one can make here is that it is increasing for $x\in\mathbb{Z}$. I have tried to make a proof but I'm not sure if it is valid. Here it goes.
Say $x,y\in\mathbb{Q}$. Then they can be written $x=\frac{p}{q},y=\frac{m}{n}$, with $q,n>0$. Because $$\frac{p}{q}=x<y=\frac{m}{n},$$
it must also hold that $pn<qm$. This implies that $a^{pn}<a^{qm}$. This would imply that $a^{\frac{p}{q}}<a^{\frac{m}{n}}$, QED.
It feels like the last step, namely "This would imply that $a^{\frac{p}{q}}<a^{\frac{m}{n}}$" in this proof depends on the fact we're trying to prove. Am I correct? If so, any suggestions as to what approach I should use to prove this theorem?
You can just do a proof by contradiction... if $a^{p \over q} \geq a^{m \over n}$, then use the monotonicity of $x \rightarrow x^{qn}$ to get that $a^{pn} \geq a^{mq}$, a contradiction. Don't even have to worry about inverses of monotone functions.