Are there infinitely many nonzero Euler quotients $a(n)=\frac{2^{\phi(n)}-1}{n} \bmod n$?

Question

This might be related to an open problem.

For odd natural $n$ define the Euler quotient:

$$ a(n)=\frac{(2^{\phi(n)}-1) \bmod n^2}{n}=\frac{2^{\phi(n)}-1}{n} \bmod n$$

Q1 Are there infinitely many $n$ with unbounded smallest prime factor for which $a(n)$ is non-zero?

Q2 Are there infinitely many $n$ for which $a(n)$ is non-zero?

$a(n)=0$ is OEIS Wieferich numbers (1)

$a(n)=1$ for prime $n$ is OEIS A125854

$a(n)=3$ for prime $n$ is OEIS A175866

Conditional results are welcome, but we believe abc implies positive answers.

Answer 1

Let $p$ be any odd prime. Suppose that $2^{p-1}-1=p^ku$ for some integer $u$ with $(p,u)=1$. Then by Lifting The Exponent Lemma for all integer $K\geq 0$ we have $$ 2^{p^K(p-1)}-1=p^{K+k}u_K $$ For some integer $u_K$ such that $(p,u_K)=1$. Therefore, for all $K>k$ the number $n=p^{K+1}$ satisfies $$ 2^{\varphi(n)}-1\not\equiv 0\pmod {n^2} $$