Can $ x^2-4a^n$ be a square? Let $ x,y,a,n \in {Z}>1$, Does the diophantine equation $ x^2-4a^n=y^2$ have any solutions? or How can I solve this?

Sure, let $x=a^n+1$ and $y=a^n-1$. There are many other solutions if $n$ is large and/or $a$ has many factors.  
One can get all solutions in integers $x,y$ positive, negative, or $0$ by letting $x+y=2u$, $x-y=2v$ where $uv=a^n$, and solving for $x$ and $y$.
