We first prove the following two lemmas.

Lemma. Let $p$ be a prime number and $b > 1$ be an integer such that $\gcd(p, b) = 1$. Then, for all positive integers $n$;
$$\nu_p(b^n - 1) \leq \nu_p(n) + C,$$
for some constant positive integer $C$.
Proof. Let $d$ be the order of $b$ modulo $p$ then if $n$ is not divisible by $d$ then $\nu_p(b^n - 1) = 0$ and $C = 1$ works. On the other hand, if $n$ is divisible by $d$ then for all odd $p$;
$$\nu_p(b^n - 1) = \nu_p(b^d - 1) + \nu_p(n) - \nu_p(d) < \nu_p(b^d - 1) + \nu_p(n).$$
and $C = \nu_p(b^d - 1)$ has the desired property.
Finally, if $p = 2$;
$$\nu_2(b^n-1) = \nu_2(b^d-1)+\nu_2(b^d+1)+\nu_p(n)-\nu_p(d) < \nu_2(b^d-1)+\nu_2(b^d+1)+\nu_p(n).$$
Letting $C = \nu_2(b^d - 1) + \nu_2(b^d + 1)$. Hence, $C = \max\{1, \nu_p(b^d - 1), \nu_2(b^d - 1) + \nu_2(b^d + 1)\}$. This completes our proof.

Lemma 1. Let $p$ be a prime number and $a > 1, m > 1$ be positive integers. Every prime divisor $q$ of $N_m = \frac{a^{p^m}-1}{a^{p^m-1}-1}$ is either $p$ or of the form $p^m k + 1$. In the former case $a \equiv 1 \pmod{p}$.
Proof. Let $c = a^{p^{m-1}}$ then $N_m = \frac{c^{p}-1}{c-1}$. Then, if $q$ divides $N_m$ then the order of $c$ mod $q$ is either $p$ or $1$. In the former case, $c^p = a^{p^m} \equiv 1 \pmod q$. Now, the order of $a$ mod $q$ divides $p^m$ and hence is a power of $p$. But if it is not equal to $p^m$ then $c \equiv 1 \pmod q$, contradicting to our assumption. Hence, the order is $p^m$ and it must divide $q-1$. In the latter case, $c \equiv 1 \pmod q$. Hence, $N_m = 1+c+\cdots+c^{p-1} \equiv p \equiv 0 \pmod q$. Thus $p=q$. Further, it can only happen whenever, $c = a^{p^{m-1}} \equiv 1 \pmod q$. Since $p=q$ it follows that $a^{p^{m-1}} \equiv a \equiv 1 \pmod p$, as desired.

Back to our problem. Take a prime $p$ such that $p > b(a-1)$ we prove that $n = p^r$ satisfies the statement of the problem for all large $r$. Indeed,
$$a^{p^r} - 1 = (a^p - 1) \left( \frac{a^{p^2} - 1}{a^p - 1} \right) \cdots \left( \frac{a^{p^r} - 1}{a^{p^{r-1}} - 1} \right) = (a^p - 1) N_2 \cdots N_r.$$
By the choice of $p$ there are pair-wise distinct prime numbers $q_2, \dots, q_r$ such that $q_i$ divides $N_i$, $i = 2, \dots, r$. Furthermore, $q_2 \dots q_r$ divides $a^{p^r} - 1$ hence, $(q_2 - 1) \dots (q_r - 1)$ divides $\varphi(a^{p^r} - 1)$. Since $q_i - 1$ is divisible by $p^i$, $i = 1, \dots, r$ it follows that
$$\nu_p \left( \varphi (a^{p^r} - 1) \right) \geq 2 + 3 + \dots + r \geq \frac{1}{2} r^2.$$

Taking $s = k - t$, this implies that $b^s - 1$ must be divisible by $p^{\frac{1}{2}r^2}$. By our first lemma, $\nu_p(b^s - 1) < s + C$. Hence,
$$s > p^{\frac{1}{2}r^2 - C}.$$
On the other hand,
$$a^{p^r} - 1 > \varphi(a^{p^r} - 1) = b^k - b^t = b^t(b^s - 1) > b^s - 1 > b^{p^2 \frac{1}{2}r^2 - C} - 1.$$
That is,
$$a^{p^r} > b^{p^2 \frac{1}{2}r^2 - C},$$
It is clear that for all large enough $r$, the right side will emulate the left side. Thus, it is enough to take $n = p^r$ for all large enough $r$ to ensure that there are no such $k, t$.