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TuZo wrote:

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It's not only that.

We have $(a,n)=(2,2),(2,1)$ and $(a,n)=(2^k+1,1)$, where $k\ge 0$ arbitrary. First, if $n=1$ then $2a-2=2^k$ for $k\ge 1$, forcing $a=2^{k-1}+1$. Let $n\ge 2$ in the remainder. Notice that $a^n+a-2\equiv 0\pmod{a-1}$ and that \[ a^n + a - 2 = (a-1)(a^{n-1}+a^{n-2}+\cdots+a+2). \]If $a-1=1$ then $a=2$, yielding the solutions above. So, let $a=2^u+1$ for $u\ge 1$. Notice that $a^{n-1}+\cdots+a+2$ must therefore be even, which is possible only when $n$ is even. Next, letting $a=2^u+1$ in $a^n+a-2$, we find \[ a^n + a - 2= \sum_{i\ge 2}\binom{n}{i}2^{ui} + (n+1)2^u. \]Note that the summation is divisible by $2^{2u}$, whereas $v_2((n+1)2^u) = u$, as $n+1$ is odd. So, this quantity can never be a power of 2, completing the solution.