Given a positive integer $m$, a prime $p$ and an integer $a$, I would like to prove that $$ x \equiv ay \pmod{p^{2m}} \qquad \lvert x \rvert,\lvert y \rvert \leq p^m $$ always has at least one solution $(x,y)\in\mathbb{Z^2}$ with $x,y$ not both $0$.
I have no problem if $p\mid a$. Indeed: if $\lvert a \rvert \leq p^m$ then $(a,1)$ is a solution, otherwise $p^m\mid a$ and $(0,p^m)$ will do.
If $p\nmid a$ then $a$ is invertible modulo $p^{2m}$, hence $x \equiv ay \pmod{p^{2m}}$ has many non-trivial solution. However, I don't know how to find one with $\lvert x \rvert,\lvert y \rvert \leq p^m$.
I think this follows from something resembling the pigeon hole principle as follows.
Fix $a$. We can assume that $p^m\nmid a$ as the case $p^m\mid a$ is easy. Consider the set of smallest non-negative remainders modulo $p^{2m}$ $$ S_a=\{ay\bmod p^{2m}\mid 1\le y\le p^m\}. $$ Observe that there are no repetitions, so $|S_a|=p^m$.
Let us place these $p^m$ pigeons into $p^m$ holes with hole #$j$ consisting of the interval $I_j:=[jp^m, (j+1)p^m-1]$, $0\le j<p^m$. There are two possibilities. Either two or more pigeons try to enter the same hole, or all the holes are occupied by a single pigeon.
If there is no double occupancy, then one of the pigeons flew into the interval $I_0$. If this pigeon carried the coordinate tag $y_0$, this means that the remainder of $ay_0$ is in the range $I_0$, and we have found our solution: the matching $x_0$ is the remainder of $ay_0$ modulo $p^{2m}$.
The other case is that two pigeons, $ay_1$ and $ay_2$, are in the same hole. Without loss of generality we can assume that the remainder of $ay_1$ is smaller than that of $ay_2$. Thus the remainder $r$ of $a(y_2-y_1)$ mod $p^{2m}$ is in the range $0<r<p^m$, and as $0<|y_2-y_1|<p^m$ we have, again, found a solution.