Is $n=1$ the only solution to $\operatorname{rev}\left(\sum_{i=0}^{p_n} p_{n+1}^i\right) = \sum_{i=0}^{p_n} p_{n+2}^i$?

Let $p_k$ denote the $k$-th prime number, and let $\operatorname{rev}(x)$ denote the decimal digit-reversal of a positive integer $x$. Define the consecutive-prime geometric sums: $$A_n = \sum_{i=0}^{p_n} p_{n+1}^i = \frac{p_{n+1}^{p_n + 1} - 1}{p_{n+1} - 1}, \qquad B_n = \sum_{i=0}^{p_n} p_{n+2}^i = \frac{p_{n+2}^{p_n + 1} - 1}{p_{n+2} - 1}$$

At $n=1$, where $(p_1, p_2, p_3) = (2, 3, 5)$, we get $A_1 = 1+3+9 = 13$ and $B_1 = 1+5+25 = 31$. Here we see that $\operatorname{rev}(A_1) = B_1$. (As a neat side note, both $13$ and $31$ happen to be prime—the 1st and 3rd emirps.)

I am trying to determine whether $n=1$ is the only solution to: $$\operatorname{rev}(A_n) = B_n \tag{1}$$

Since decimal reversal depends on base-10 representation rather than pure algebra, finding a simple algebraic contradiction seems difficult. However, in checking candidates computationally up to $n = 2000$, I noticed that three simple constraints act as an effective sieve that eliminates every case with $n > 1$:

1. Digit-length bounds For $(1)$ to hold, $A_n$ and $B_n$ must have the exact same number of decimal digits. Since $p_{n+2} > p_{n+1}$, the ratio grows roughly as: $$\frac{B_n}{A_n} \approx \left(\frac{p_{n+2}}{p_{n+1}}\right)^{p_n}$$ For most indices, $B_n$ has strictly more digits than $A_n$. In fact, checking all $n \le 2000$ reveals an interesting pattern: every single prime triple with a gap $p_{n+2} - p_{n+1} \ge 4$ already fails the digit-length test. Only 49 candidates survive this filter up to $n=2000$, and all 49 correspond to twin-prime gaps where $p_{n+2} = p_{n+1} + 2$.

2. Modulo 9 congruence In base 10, any integer is congruent to its reversal modulo 9, so a valid solution requires $A_n \equiv B_n \pmod 9$. Checking this congruence against our 49 surviving twin-prime candidates eliminates 42 of them, leaving just 7 candidates up to $n=2000$: $n \in \{1, 108, 600, 607, 699, 1304, 1703\}$.

3. Modulo 11 congruence Decimal reversal also interacts predictably with modulo 11: if an integer $X$ has $d$ digits, then $\operatorname{rev}(X) \equiv (-1)^{d-1} X \pmod{11}$. Thus, any solution to $(1)$ must satisfy: $$B_n \equiv (-1)^{\lfloor \log_{10} A_n \rfloor} A_n \pmod{11}$$ When applying this condition to the 7 remaining candidates, all 6 cases with $n > 1$ fail. For example, at $n=108$ (the primes $(593, 599, 601)$), both $A_{108}$ and $B_{108}$ have $1,648$ digits and pass mod 9, but $B_{108} \equiv 4 \pmod{11}$ whereas $\operatorname{rev}(A_{108}) \equiv 9 \pmod{11}$.

Questions

1. Can it be rigorously proven that no solutions exist for $n > 1$? Specifically, can Baker's theorem on linear forms in logarithms or analytic prime-gap bounds be used to prove that $\lfloor \log_{10} A_n \rfloor < \lfloor \log_{10} B_n \rfloor$ for all sufficiently large $n$ when $p_{n+2} - p_{n+1} \ge 4$?
2. More generally, if we drop the requirement that the bases must be consecutive primes, are there any known finiteness results for the base-10 repunit reversal equation: $$\operatorname{rev}\left(\frac{a^m - 1}{a - 1}\right) = \frac{b^m - 1}{b - 1}$$ for $m \ge 3$ and $a, b \ge 2$? I would appreciate any references to literature on Diophantine equations involving digit reversals.