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Let $2^k$ be the single power of two on the interval $[n, 2n-2] \implies n \le 2^k \le 2n-2$. Let's prove that $a^{2^k}+1$ satisfies the condition. SFTSoC it doesn't, that means there must exist a odd prime $p$ and $n \le m \le 2n-2$ with $m \neq 2^k$ such that $p|a^{2^k}+1$ and $p|a^m+1 \implies a^{2^k} \equiv a^m \equiv -1 \pmod{p}$, call $ord_pa=t \implies t \nmid 2^k$ (because $a^{2^k} \equiv -1 \not\equiv 1 \pmod{p}$ as $p \neq 2$), but $a^{2^{k+1}} \equiv a^{2m} \equiv 1 \pmod{p} \implies t|2^{k+1} \implies t=2^{k+1}$ (since $t \nmid 2^k$) so $t=2^{k+1}|2m \implies 2^k|m$, contradiction! Since $2^k$ is the biggest power of two in that interval and $m \neq 2^k$.

Here we go with the generalisation: WLOG, assume that $a$ is odd. We will show that, at least one of the numbers amongst $a^m+1, a^{m+1} \cdots, a^n+1$ is coprime with all other terms. Let $p, e$ be two numbers such that: $m \ge p, q \ge n$ with $d=(p, q)$. Let $dp'=p, dq'=q$. Note the following statement: At least one of $p', q'$ is even if and only if $a^d+1$ divides $a^p+1, a^q+1$. Thus, taking $\nu_2$ of the exponents: It suffice to find a unique number amongst: $\nu_2(m), \nu_2(m+1) \cdots \nu_2(n)$. Note that, the maximum amongst the numbers should be unique. Suppose if not: consider two numbers $r=2^k l_1, s = 2^k l_2$ such that $k$ is the maximum number amongst them in the sequence. WLOG assume $s>r$. \ Thus, since $l_1, l_2$ are odd integers, there exists an even integer between them. Consider a number $l'$ which is even: then: $\nu_2(2^k l') \ge k+1$ along with $s > 2^k l' > r$ which leads to contradiction.