$$(43)_x=(y3)_8$$
Since a number in base$-k$ can only have digits from $0$ to $(k-1)$, we can conclude that: $x \geq 5$ and $y \leq 7$
Now, the original equation, when converted to decimal base gives:
$$\begin{align}4x^1 + 3x^0 &= y(8^1)+3(8^0)\\[0.8em]4x + 3 &= 8y + 3\\[0.8em]x&=2y\end{align}$$
So, we have the following constraints:
$$x\geq 5\\y \leq 7\\x=2y\\x,y \text{ are integers}$$
The set of values of $(x,y)$ that satisfy these constraints are:
$$\begin{align}(x&,y)\\\hline\\(6&,3)\\(8&,4)\\(10&,5)\\(12&,6)\\(14&,7)\end{align}$$
I am counting 5 pairs of values.