An alternative method is to exploit the rotational invariance of the Gaussian. You can write
$$X^T Y = |X| \left( \left(\frac{X}{|X|}\right)^T Y \right).$$
Because $Y$ is rotationally invariant, the inner product is now independent of $X$, and in fact just has distribution $N(0,1/m)$. Now let $C>1$ be an arbitrary parameter. We can bound the probability $X^T Y > \epsilon$ by the probability one of the following two events occur.

$ \left(\frac{X}{|X|}\right)^T Y \geq \frac{\epsilon}{C}$. Assuming $ \epsilon \sqrt{m}/C$ tends to infinity, this occurs with probability $\Phi (\frac{\epsilon \sqrt{m}}{C})=(1+o(1)) \sqrt{\frac{m}{2 \pi}} \exp(-\frac{\epsilon^2 m}{C^2})$.

$|X| \geq C$. The norm of a Gaussian vector is well studied, and it is standard (see, for example Chapter 2 of these notes, that $|X|$ is tightly concentrated around its expectation. For example, applying Corollary 2.3 of the linked notes gives that the probability this occurs is at most $\exp(-\frac{1}{4} (1-\frac{1}{C^2})^2 m)$

For $\epsilon$ bounded away from $0$ you can choose $C$ to optimize the sum of the two terms getting a bound that is exponential in $m$ but with a non-optimal exponent. If $\epsilon$ is tending to $0$ with $m$, then the first term is dominant. That term remains small so long as $\epsilon$ is much larger than $\sqrt{\frac{\log m}{m}}$.