I am having difficulty solving the above problem.
I understand that the number of strings which have ONLY an $x$ in it is $26^8 - 25^8$, and the same goes for $y$, but I cannot wrap my head around how to solve for $x$ and $y$.
The answer I came up with is ${8}\choose{2}$ $\times 26^6$, the $8\choose 2$ coming from placing $x$ and $y$ into any of the $8$ spots onto the strong, and the $26^6$ filling out the rest of the spots with any letters.
I cannot shake the feeling that this answer is incorrect, and I am looking for someone to help walk me through the problem so I can get a better understanding of it. Thanks.
Your answer is indeed incorrect because it has overcounted scenarios where you have more than one $x$ and/or more than one $y$ occurring. For example, letting the positions chosen in the first step be colored red, you would have considered $\color{red}{XY}xaaaaa$ to be a different outcome than $x\color{red}{YX}aaaaa$ despite them both having the same letters in the same order.
For a corrected approach, consider counting the opposite event: We have no $X$'s or we have no $Y$'s.
Let $A$ be the set of 8 letter strings where we have no $X$'s. Let $B$ be the set of 8 letter strings where we have no $Y$'s. Let $\Omega$ be the set of 8 letter strings with no restrictions.
Remember that $|A\cup B| = |A|+|B|-|A\cap B|$ and remember that $|\Omega| - |A\cup B| = |A^c\cap B^c|$, the total you are really looking for.