When $p$ is a small prime, finding a solution to $a^x \equiv b \bmod p$ is correspondingly easy (assuming that $a,b$ are not divisible by $p$). For example, $2^x \equiv 28 \bmod 3$ holds for any even power $x$.
It is also clear that if $a^x \equiv b \bmod p^s$ for some integer power $s\ge 1$, then:
$$ a^x \equiv b \bmod p^s \implies a^x \equiv b \bmod p^{s-1} \implies \ldots \implies a^x \equiv b \bmod p $$
Eric Bach (1984) wrote up a polynomial time algorithm, similar to the ideas of Hensel lifting for polynomials, to work backwards in that chain of implications. His purpose was to compare the difficulty of solving the discrete logarithm problem for a general modulus $n$ with that of factoring $n$, so we here are concerned only with a portion of his work.
As usual $\mathbb{Z}$ is the integers, and $\mathbb{Z}_n$ will denote the ring of integers modulo $n$. The additive group of $\mathbb{Z}_n$ will be written as $\mathbb{Z}_n^+$, while the multiplicative group of its units is written $\mathbb{Z}_n^*$. Note that the additive group is cyclic and has $n$ elements, while the number of elements in the multiplicative group is given by Euler's totient function $\varphi(n)$ (counting the integers $0\lt k \lt n$ which are coprime to $n$).
Taking the case where $n = p^s$ is a prime power $s\ge 1$, the counting is easy:
$$ \varphi(p^s) = (p-1)p^{s-1} $$
Is the multiplicative group $\mathbb{Z}_{p^s}^*$ cyclic of that order? In the case that $p$ is an odd prime, the answer is yes, and a generator for this group is called a primitive root modulo n.
For this reason Bach begins with the case of an odd prime $p$, and thus we know an isomorphism:
$$ \mathbb{Z}_{p^s}^* \cong \mathbb{Z}_p^* \times \mathbb{Z}_{p^{s-1}}^+ $$
It is easy to find the projection of $k \in \mathbb{Z}_{p^s}^*$ onto the first factor: just take the reduction of $k \bmod p$.
It is the essence of the rest of the algorithm to show a polynomial-time computable projection $\theta$ from multiplicative group $\mathbb{Z}_{p^s}^*$ homomorphically onto the second factor, additive group $\mathbb{Z}_{p^{s-1}}^+$:
$$ \theta: \mathbb{Z}_{p^s}^* \to \mathbb{Z}_{p^{s-1}}^+ $$
Once we have this function, the "lifting" of a solution mod $p$:
$$ a^{x_1} \equiv b \bmod p $$
to one in $\mathbb{Z}_{p^s}^*$:
$$ a^{x_s} \equiv b \bmod p^s $$
proceeds in a fairly natural way. Note that $x_s \equiv x_1 \bmod (p-1)$.
Applying the projection $\theta$ to the above "higher order" congruence leads us to this problem in the additive group $\mathbb{Z}_{p^{s-1}}^+$:
$$ c \cdot \theta(a) \equiv \theta(b) \bmod p^{s-1} $$
which can be solved by reciprocating $\theta(a)$ in $\mathbb{Z}_{p^{s-1}}$ via the extended Euclidean algorithm.
If we find an integer $x_s$ by Chinese remainder theorem such that both:
$$ x_s \equiv x_1 \bmod (p-1) $$
$$ x_s \equiv c \bmod p^{s-1} $$
then we are done, because by the isomorphism:
$$ a^{x_s} \equiv b \bmod p^s $$
Bach gives this formula for the homomorphism $\theta:\mathbb{Z}_{p^s}^*\to \mathbb{Z}_{p^{s-1}}^+$:
$$ \theta(k) = \left[ \frac{k^{(p-1)p^{s-1}} - 1}{p^s} \right] \bmod p^{s-1} $$
Details of why this is a homomorphism from the multiplicative group to the additive one are given in Bach's paper. Certainly $\theta(1) = 0$, and as I find time, I'll add further proof.
As Bach notes, the complexity of this function is polynomial in $\log p$ and $s$ if the numerator (of the integer expression inside the square brackets) is evaluated mod $p^{2s-1}$. We need that many radix $p$ places to get the right residue mod $p^{s-1}$ after dividing by $p^s$. The exactness of the division is implied by Euler's generalization of Fermat's Little Theorem.
Example (taken from the Question)
Consider the equation $2^x \equiv 28 \bmod 3^5$. As already remarked above, any even integer $x$ solves the "base case" $2^x \equiv 28 \bmod 3$. Therefore we will get a second congruence to combine with the first one:
$$ x \equiv 0 \bmod (3-1) $$
$$ x \cdot \theta(2) \equiv \theta(28) \bmod 3^4 $$
Recall that computing $\theta(k)$ involves raising to the power $\varphi(3^5) = 2\cdot 3^4 = 162$. A minimal additive chain for exponents leading to $162$ has nine steps. The binary one attains this by repeated squaring, doubling exponent to $128$, followed by multiplications of intermediate values that add to the exponent:
$$ 162 = 128 + 32 + 2 $$
In any case the values are small enough to do by calculator and to check with a spreadsheet:
$$ \theta(2) \equiv \left[ \frac{2^{162}-1}{243} \right] \equiv 16 \bmod 81 $$
$$ \theta(28) \equiv \left[ \frac{28^{162}-1}{243} \right] \equiv 18 \bmod 81 $$
Thus the second congruence amounts to:
$$ 16x \equiv 18 \bmod 81 $$
$$ 80x \equiv 90 \bmod 81 $$
$$ -x \equiv 9 \bmod 81 $$
$$ x \equiv 72 \bmod 81 $$
Since $x=72$ is already even, this is our solution modulo $2\cdot 3^4$.
TO DO: Fill in the details for the exceptional "even prime" case $p=2$.
