Ok, new try (after the first was spam): Let $p_i$ be the smallest prime dividing $n_i$ ($n_i=1$ is not possible because $1|a-1$). Claim: $p_i|a-1$, especially $p_i < n_i$ (otherwise $n_i=p_i|a-1$). Assume the contrary: Let $q_i$ be any prime dividing the order of $a \mod p_i$ (the order is $>1$ if $p_i \nmid a-1$). As we have $p_i|n_i|a^{n_i}-1$ and $p_i|a^{p_i-1}-1$, we get $q_i|n_i, (p_i-1)$, contradiction since $q_i<p_i$ divides $n_i$. For each $i$ we can choose one number from $c_i \in \{1,p_i,n_i\}$ and look at $m=\prod_{i=1}^k c_i$. We have $a^{c_i} \equiv 1 \mod c_i$ independet what $c_i$ we took, thus also $a^{m} = \left( a^{c_i} \right)^\frac{m}{c_i} \equiv 1 \mod c_i$. But the $c_i$ are coprime, giving $a^m \equiv 1 \mod m$ and we always get different $m$ by different choices of $c_i$. In total we got a total number of $3^k$ such $m$ (where $m=1$ was counted). But the inequality $3^k-1 \geq 2 \cdot 2^k-2$ is trivial to check to be true.

I will post an alternative solution. proposition：Let $c$ and $h$ be positive integers. If $a^{c^{h}}-1$ is divisible by $c^{h}$,then $a^{c^{h+1}}-1$ is divisible by $c^{h+1}$. proof：Obviously, $a^{c^{h}} \equiv 1\pmod{c}$. $(a^{c^{h}})^{c-1}+(a^{c^{h}})^{c-2}+\cdots+1 \equiv 1+1+\cdots+1 \equiv c \equiv 0 \pmod{c}$. Therefore, $a^{c^{h+1}}-1 = (a^{c^{h}}-1)( (a^{c^{h}})^{c-1}+(a^{c^{h}})^{c-2}+\cdots+1) $ is divisible by $c^{h+1}$. Now back to the problem. This proposition implies that a positive integer $b$ such that $b>1$ if $a^{b} \equiv 1 \pmod{b}$ then $a^{b^{t}} \equiv 1 \pmod{b^{t}}$ for an arbitrary positive integer $t$. It implies that infinity many positive integers $x$ satisfy $a^{x} \equiv 1 \pmod{x}$, of course, more than $2^{k+1} -2$.