You have $3$ possibilities:
$$x^3=x^5 \Rightarrow x^2=e \,.$$
In this case since $x^{15}=e$ it follows that $x=e$, which is not possible (since you only get one value in your set).
$$x^5=x^9 \Rightarrow x^4=e \,.$$
Again, this implies that $x=e$, not possible.
$$x^3=x^9 \Rightarrow x^6=e \,.$$
Thus, $x^3=x^{\operatorname{gcd}(6,15)}=e$. This means that $x$ must have order $1$ or $3$, but again $x=e$ is not possible.
Thus, $x$ is an element of order $3$ in your group, and from there it is easy: $x^{13n}=x^{13m} \Leftrightarrow 3|13(m-n) \Leftrightarrow 3|m-n$... So how many distinct values do you get?
List the numbers that give remainder $1$ on division by $5$, looking for remainder $2$ on division by $3$. Start listing: $1$, $6$, $11$. Bingo!
Remark: This is probably the quickest way to solve the GRE problem. An alternative is to list the numbers that leave remainder $2$ on division by $3$, and look for remainder $1$ on division by $5$. That is a little slower: things increase faster if we count by $5$'s than if we count by $3$'s.
As to elegance, that requires thinking, and the GRE is not about that. On to the next question.
Best Answer
Our expression $2^{14}\cdot 3^{24}$ can be rewritten as: $$2\cdot 3\cdot 2^{13}\cdot 3^{23}$$ $$6\cdot 2^{13}\cdot 3^{23}$$ To make another $6$ factor, we just borrow a $2$ and a $3$ like this: $$6^2\cdot 2^{12}\cdot 3^{22}$$ How many $6$s can we make? We can make as many $6$s as we want until either $2$ or $3$ runs out. In this case we can make as many $6$s as we want until $2$ runs out. $$6^{14}\cdot 3^{10}$$ Therefore $y=14$