Let $(x_1,y_1)$ and $(x_2,y_2)$ be two different solutions to $ax^2+by^2=mp^r$. We try to deduce some lower bounds on the solutions, in order to reach some contradiction. Observe that $$(x_1y_2)^2\equiv (x_2y_1)^2\pmod{p^r}$$ thus $p^r\ |\ (x_1y_2-x_2y_1)(x_1y_2+x_2y_1)$. Note that none of $x_i,y_i$ can be a multiple of $p$, so $x_1y_2-x_2y_1,x_1y_2+x_2y_1$ cannot be multiples of $p$ simultaneously. Therefore one of them must be a multiple of $p^r$. **Case 1:** $p^r | x_1y_2+x_2y_1$. Then $x_1y_2+x_2y_1\geq p^r$, then by a chain of inequalities $$(ax_1^2+by_2^2)+(ax_2^2+by_1^2)\geq 2\sqrt{ab}x_1y_2+2\sqrt{ab} x_2y_1 > 2mp^r$$ and a contradiction occur since the LHS and RHS above should be equal. **Case 2:** $p^r | x_1y_2-x_2y_1$. Some quick discussion on the coprimality of the solutions imply that $x_1y_2\neq x_2y_1$. WLOG assume $x_1y_2 > x_2y_1$. Then $x_1y_2 > p^r$, and by similar methods to one used in Case 1 we deduce $$(ax_1^2+by_2^2)\geq 2\sqrt{ab}x_1y_2 > 2mp^r$$ and again a contradiction occur since the LHS must be smaller than RHS.

An alternative approach to avoid the case-work is to conclude that $x_1y_2 + x_2y_1 \geq p^r$ in both cases. Seems better.