Any solution yet? bump

Suppose $q^c = 1 + p + \ldots + p^m = \frac{p^{m+1} - 1}{p - 1}$. Firstly suppose $m + 1$ is not prime, say $m + 1 = ab$, then $\frac{p^a - 1}{p - 1}$ divides $q^c$, and hence must be a power of $q$. However, by Zsigmondy's theorem, unless $(m + 1, p) = (6, 2)$ or $(2, 2^t - 1)$ for some $t \ge 2$, this is impossible. Case 1: $(m + 1, p) = (6, 2)$: in this case, we have $(m, p) = (5, 2)$, so $q^c = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 = 63$, which is impossible. Case 2: $(m + 1, p) = (2, 2^t - 1)$ for $t \ge 2$: in this case, we have $(m, p) = (1, 2^t - 1)$, so $q^c = 1 + 2^t - 1$, so $q = 2$, which is impossible as $p < q$. Doing the same thing by swapping the roles of $p, q$ gives us a family of solutions $(p, q) = (2, 2^t - 1)$ for some $t \ge 2$. Note that $t$ must be prime as otherwise, for any proper divisor $d \ge 1$ of $t$, $2^d - 1$ divides $q$, which is impossible. So now suppose that both $m + 1$ and $n + 1$ are prime. Suppose $p^d = \frac{q^{n + 1} - 1}{q - 1}$. Then we have $q \mid p^d - 1$, so $q \mid \gcd(p^d - 1, p^{m + 1} - 1) = p^{\gcd(d, m + 1)} - 1$. Now since $m + 1$ is prime, we must have $m + 1 \mid d$, as otherwise $q \mid p - 1$, which is impossible. So we must have $q^c \mid p^{m + 1} - 1 \mid p^d - 1 = \frac{q^n - 1}{q - 1} \cdot q$, so $c = 1$. Now note that $q - 1 = \frac{p^m - 1}{p - 1} \cdot p$, so $p \mid q - 1$. If $p = 2$, then we get to the desired solution. Now suppose $p > 2$. Then by lifting the exponent lemma, we have $d = \nu_p \left(\frac{q^{n + 1} - 1}{q - 1}\right) = \nu_p(n + 1)$. Since $n + 1$ is prime, and $d \ne 0$, we must have $n + 1 = p$ and $d = 1$. However, this leads to a contradiction as $p < q$.