integer solutions to $a^m+nx^2 = y^n$ with various conditions

I consider the following equation with conditions of obtaining solutions

$$a^m+nx^2 = y^n$$

This equation has solution when $a$ is an even prime and $x, y, m$ are positive integers with $(nx, y) = 1$ and $n>1$.

If we fix $a$ as an odd prime and without restriction on $x, y, m$ and $n$ (may be odd or even) with $n > 1$ and $(nx, y) =1$, can we have solutions?
If yes, how to determine such solutions?
If there is empty solution, how to conclude?

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For $n=2$, you have a Pellian equation $y^2-2x^2=a^m$. If $a\equiv\pm1\pmod8$ then $y^2-2x^2=a$ has infinitely many solutions, as does $y^2-2x^2=a^m$ for each $m$. The easiest way to see this is via the ring of integers in the field ${\bf Q}(\sqrt2)$. – Gerry Myerson Sep 28 '12 at 7:30
@Gerry Myerson! if n= 3 and n = 5, what are the solutions? How can we obtain? Thank you for the reduced pellian equation. – VMRFDU Sep 28 '12 at 8:29
How much do you know about elliptic curves? – Gerry Myerson Sep 28 '12 at 9:24
I'm sorry, but explaining elliptic curves takes a lot more than a comment and a graph. Get yourself a book like Silverman's and start reading. But first make sure you know your elementary Number Theory, your Algebraic Geometry, your Commutative Algebra, and your Complex Analysis, as you will need them. – Gerry Myerson Sep 29 '12 at 6:52
For $n=3$, you can rewrite the equation as $X^2=Y^3-27a^m$ (where $X=9x$ and $Y=3y$). This is called a Mordell equation. There is no general way to tell how many integer solutions it has for given $a$ and $m$, or even whether it has any at all. There is an enormous amount of literature about it, which I encourage you to search. – Gerry Myerson Oct 1 '12 at 7:09