Let $x$ and $q$ be an integer.Also, $a$ and $b$ are integers.
We know the two modulus equations.
i) $x \equiv y$ mod $p^a -1$
ii) $x \equiv z$ mod $p^b -1$
Then how to find $x$? Can we find $x$ exactly? or Can we find the candidates of $x$? If then how many candidates of $x$?
Maybe we do not have any $x$, but if we have one, then we have many. Let $h = (p^a -1)(p^b-1)+1$. Then we have \begin{align} h \equiv 1 \mod p^a-1 \\ h \equiv 1 \mod p^b-1 \end{align} and if $x$ be a solution, then $hx$ is a solution too.
If $a \mid b$ or $b \mid a$ then $y$ and $z$ may be such that we do not have any choice for $x$. For example if $p = 2,a =2,b=4,y =2$ and $z = 0$ then there is no $x$ to fit in both equations.
In fact if we can find $x_1$ and $x_2$ such that \begin{equation} (p^a - 1)x_1 \equiv z \mod p^b-1\\ (p^b - 1)x_2 \equiv y \mod p^a-1 \end{equation} then $x = (p^a - 1)x_1 + (p^b - 1)x_2 + k(p^a -1)(p^b-1)$ gives a family of solutions. Based on this, we can say that if $(a,b)=1$ then (as $(p^a -1,p^b-1) = (a,b) $ ) always a solution exists.
Hope this be helpful.
Sorry! $(p^a -1,p^b-1) = p^{(a,b)}-1$, and my last result may not be true.