I believe the claim in question may not hold, although it seems to be tricky to construct a counterexample.

Nevertheless, Theorem 1 is correct and represents a partial case of the generalized Pocklington primality test. In fact, rather than requiring $a<p$, it is enough to require that $a<m$ or $a<\sqrt{n}$.

Correspondingly, the given proof of Theorem 1 is easy to make work: instead of concluding that $m\mid\phi(n)$ and relying on the unproved claim, one can show that $m\mid (r-1)$ for every prime divisor $r\mid n$, implying that $n$ does not have prime divisors below $\sqrt{n}$ and thus it must be prime.

I believe the claim in question may not hold, although it seems to be tricky to construct a counterexample.

Nevertheless, under the replacement of $b_i^{(n-1)/q_i}\not\equiv 1\pmod{n}$ with $\gcd{(b_i^{(n - 1)/q_i} - 1, n)} = 1$, Theorem 1 is correct and represents a partial case of the generalized Pocklington primality test. In fact, here rather than requiring $a<p$, it is enough to require that $a<m$ or $a<\sqrt{n}$.

From practical perspective, if it happens that $b_i^{(n-1)/q_i}\not\equiv 1\pmod{n}$ but $\gcd{(b_i^{(n - 1)/q_i} - 1, n)} > 1$ then this gcd gives a non-trivial divisor of $n$.

Correspondingly, the given proof of Theorem 1 is easy to make work: instead of concluding that $m\mid\phi(n)$ and relying on the unproved claim, one can show that $m\mid (r-1)$ for every prime divisor $r\mid n$, implying that $n$ does not have prime divisors below $\sqrt{n}$ and thus it must be prime.