The Prime $x^{y^2+2y-8}-1$

First we shall show that for a positive integers $n,a$ with $n>1$ , $a^n-1$ is prime only when $a=2$ .

Now it is clear that $a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$

So for $a>3$ it is clear that $a^n-1$ is composite. For $a=1$ it is clearly not true. And if $a=2$ , $a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)=a^{n-1}+a^{n-2}+...+a+1$ , which is possibly prime. [ $n=2,3$ for example.]

Next we show that $n$ has to be prime.

Suppose $n$ is composite, $n=cd$ with $c,d>1$ . Then we have,

$a^n-1=a^{cd}-1=(a^c-1)(a^{(c-1)(d)}+a^{(n-2)(d)}+...+a^{2d}+a^d+1)$

Which is then clearly composite as both factors are larger than 1. So $n$ has to be prime too.

Now back to the question. As, $y^2+2y-8=(y-2)(y+4)$ . It is clear that $y=3,-5$ are the only $y$ values that will result in $y^2+2y-8$ being prime.(Both will result value in the value $7$ .) So letting $x=2$ ;

$2^7-1=127$

which is prime. Hence $(2,3)$ and $(2,-5)$ are solutions.

Thus far, we only considered $x>0$ and $y^2+2y-8>1$ . So,we need to consider the other cases too.

If $y^2+2y-8<0$ , then $x^{y^2+2y-8}-1$ would not be an integer. If $y^2+2y-8=0$ , then the result of $x^{y^2+2y-8}-1$ would be $0$ or undefined. Also it is clear that $y^2+2y-8=1$ has no integer solutions. Therefore, $y^2+2y-8>1$ for all cases for $x^{y^2+2y-8}-1$ to be prime.

If $x=0$ , then $x^{y^2+2y-8}-1$ will results in $-1$ or undefined. If $x<0$ , then $x^{y^2+2y-8}-1$ will be a negative integer for an odd $y^2+2y-8$ . For an even $y^2+2y-8$ , it is the same as considering $x>0$ and $y^2+2y-8>1$ . So there are no solutions when $x<0$ as $y^2+2y-8$ needs to be prime.

Thus the only solutions are $(2,3)$ and $(2,-5)$ .