First of all, simplify to $\left\lceil \frac{\log(\left\lceil \frac{y}{x} \right\rceil)}{\log x-1} \right\rceil = 1.$ Without loss of generality, $\log$ denotes the natural logarithm. Using the definition of $\lceil \cdot \rceil$, one finds 0 < \frac{\log \left\lceil \frac{y}{x} \right\rceil}{\log x-1} \le 1.
For now, assume $x > 2, y > 1$. Then $\log x - 1 \ge \log 1 = 0$, so 0 < \log \left\lceil \frac{y}{x} \right\rceil \le \log (x-1). Taking exponentials on both sides, we get 1 < \left\lceil \frac{y}{x} \right\rceil \le x-1.
Using the definition of $\lceil \cdot \rceil$ again, we find that since 1 < \lceil y/x \rceil, 1 < y/x, so $y > x$. Also, since $\lceil y/x \rceil \le x-1$, we have $y/x \le x-1$ as well, i.e., $y \le x^2-x$. In other words, solutions with $x > 2, y > 1$ will always be of the form x < y \le x^2 - x. Edit: Not all of these tuples fulfill the equation; you will need to check for every concrete pair.
Edit: If $x > 2$, it is also easy to see that if $y \le 1$, 0 < \log\lceil y/x\rceil fails to hold, so the above lists all solutions for $x > 2$.
In the case that 1 < x \le 2, a similar calculation yields $0 > \log \lceil \frac{y}{x}\rceil \ge \log x-1$, so $1 > \left\lceil\frac{y}{x}\right\rceil \ge x-1$. By choise of $x$, we have 0 < x-1 \le \lceil y/x\rceil < 1. Since $\lceil y/x \rceil$ is an integer, no solution exists in this case.
Finally, for $x \le 1$, $\log_{x-1}$ is not defined over $\mathbb R$ anymore, so these cases can be excluded.
All in all, the solutions to this equation are of the form x > 2, x < y \le x^2-x, but not every tuple of this form is a solution.
Edit: For a more exact solution, see Henry's post.