Say we do know $b$ but not $k$, and are given $g$ such that $g\equiv b^k\pmod p$. And say there exist factors $E = e + m'p$ ($e \equiv b^i \bmod p$) and $F = f + m''p$ ($f \equiv b^j \bmod p$) of $g$. Then $EF = g + pm = b^k$ for some $m$, $m'$ and $m''$.
So if we can find the right $(m, m')$ we can factor $G = g+pm = EF$. Then $E$ is congruent modulo $p$ to a power of $b$, and so is $F$. But note that $E$ and $F$ may not be perfect powers of $b$.
So we generate values of $E$ that are perfect powers of $b$, and then see if we can factor a member of the congruence class of $g$ mod $p$ — i.e. are any members of the congruence class of $g$ mod $p$ multiples of $E$? We can use the Extended Euclidean algorithm for this if $g$ is $\pm1$. Maybe we could also use the EEA for this if $g$ is any number $d$, by reducing the $g+pm$ and $E$ by the multiplicative inverse of $g$ mod $p$ (which we can find).
So if we can find a solution in $F$ to : $g \equiv b^i F\bmod p$ then we can reduce the problem to finding the exponent (to base $b$) of $F$ mod $p$. So then shouldn't we be able to find a sequence of exponents of our perfect power (the $E$s, the exponents being the $i$s) = until $F = 1$. At that time the sum of those exponents will be congruent (mod $p$) to the exponent of $g$.
It is on this basis that I claim that factoring is equivalent to the discrete log problem.
Is this hopelessly confused, or is there something here?