Problem: Prove that if $a \geq 2$ and $a^n+1$ is a prime number, then $a=2$ and $n=2^k$ for some non-negative integer k.
First of all, I do not understand what the question is asking me. Should I prove the problem through Contradiction, Converse, or Contrapositive?
Note that $6^2+1=37$, so the assertion as stated is not true. There are many other known counterexamples, such as $10^2+1$ and $6^4+1$.
Certainly $a$ must be even, that is, divisible by $2$. And $n$ must be a power of $2$. One can prove it by showing that if $n$ has a proper odd divisor, then $a^n+1$ cannot be prime. And the most straightforward way of showing that a number $N$ is not prime is by exhibiting a non-trivial divisor of $N$.
Since you asked only about strategy, we will say no more unless asked!
Added: Suppose that $a^n+1$ is prime, with $a\ge 2$. Then $a$ cannot be odd. For if $a$ is odd, then $a^n+1$ is even and greater than $2$, so cannot be prime.
Now we show that $n$ must be a power of $2$. Suppose to the contrary that $n$ is not a power of $2$. Then $n$ has an odd divisor $d\gt 1$. Let $n=qd$. Then $$a^{n}+1=a^{qd}+1=(a^q)^d+1.$$ Now if $d$ is odd, then $x^d+1=0$ has the solution $x=-1$, so the polynomial $x+1$ divides the polynomial $x^d+1$. More explicitly, $$x^d+1=(x+1)(x^d-x^{d-1}+x^{d-2}-\cdots+1).$$ So in particular the number $a^q+1$ is a divisor of the number $(a^q)^d+1$. It is clear that $1\lt a^q+1\lt (a^q)^d+1$, so $a^q+1$ is a non-trivial divisor of $(a^q)^d+1$, that is, of $a^n+1$.
Thus if $n$ is not a power of $2$, then $a^n+1$ cannot be prime.
Remark: Instead of using factorization, we can use congruences to show that $a^q+1$ divides $(a^q)^d+1$. For let $m=a^q+1$. Then $a^q\equiv -1\pmod{m}$.
Thus since $d$ is odd, $(a^q)^d\equiv (-1)^d\equiv -1\pmod{m}$. But this is a fancy way of saying that $m$ divides $(a^q)^d+1$.
