Bulgarian Winter Tournament

Let $d = (x, y)$ be the greatest common divisor of the numbers $x$ and $y$. Then, $x = d x_1$, $y = d y_1$, where $x_1, y_1$ are coprime naturals. The expression in the condition can be rewritten as
$$\frac{d^{k-1}}{y_1^2 - x_1^2} x_1^k y_1 = p,$$
where we want $p$ to be prime. But $(x_1^k, y_1^2 - x_1^2) = (x_1, y_1)^{\min(k,2)} = 1 = (y_1, y_1^2 - x_1^2)$, i.e., $x_1^k y_1$ is coprime to $y_1^2 - x_1^2$ and for $p$ to be whole it takes $y_1^2 - x_1^2 \mid d^{k-1}$. Moreover, $y_1 > x_1$ and therefore $x_1^k y_1 > 1$. Therefore, to be $p$ it is simply necessary that $x_1 = 1$, $y_1 = p$ and $y_1^2 - x_1^2 = d^{k-1}$ are fulfilled at the same time. If $k=1$, then $1 = p^2 - 1 = (p-1)(p+1)$, which is impossible. At $k=2$ we want $d = p^2 - 1$ and here the pair $(x, y) = (p^2 - 1, p(p^2 - 1))$ has the desired

property for every prime $p$. Therefore, $k=2$ is a solution. With $k > 2$ we want
$d^{k-1} = p^2 - 1 = (p-1)(p+1)$. It is directly verified that $p=2$ does not lead to
a solution, since 3 is not the $k-1$th power of a natural number. Hence, $p$ is an
odd number and $8 \mid p^2 - 1$, whence $k-1 \ge 3$. Therefore, $k=3$ is not a solution.
$k=4$ is a solution at $p=3$ and $d=2$, i.e., $(x, y) = (2, 6)$. $\square$