Does $\operatorname{product-of-digits}(n)=n/3$ have at least one solution in any bases except for the power of $3$?

Not a full solution, but I’ll add that

• For $b = (3i - 1)j$ with $i, j \ge 1$ we have a two-digit number $n = (i)b + (ij)$.

  This narrows the potential counterexamples $b$ to those that are a product of primes of the form $3i + 1$ or $3$. This set has zero asymptotic density.

• For $b = 6i + 1$ with $i \ge 1$ we have a second three-digit number $n = (2i + 1)b^2 + (3i + 1)b + (4i + 1)$.

And here are some two-parameter three-digit families. Since $b$ is linear in $j$, these solve an entire congruence class for each fixed $i$:

• For $b = (2i - 1)(3(3i - 1)j + 1)$ with $i, j \ge 1$, we have a three-digit number $n = (1)b^2 + ((3i - 1)^2 j + i)b + (2i - 1)((9i - 4)j + 1)$.

  (Solves $b = 6j + 1, 45j + 3, 120j + 5, 231j + 7, \ldots$.)

• For $b = (3i - 2)((3i - 1)j - 2i + 1)$ with $i \ge 2, j \ge 1$, we have a three-digit number $n = (1)b^2 + ((3i - 2)^2 j - 3(i - 1)(2i - 1))b + (2i - 1)((3i - 1)j - 2i + 1)$.

  (Solves $b = 20j - 12, 56j - 35, 110j - 70, 182j - 117, \ldots$. Too bad the third digit is too big for $i = 1$—otherwise we’d get $b = 2j - 1$ and we’d be done!)

• For $b = i((9i - 3)j - 6i + 1)$ with $i, j \ge 1$ (except $i = j = 1$) we have a three-digit number $n = (1)b^2 + ((3i - 1)^2 j - 3i(2i - 1))b + i((6i - 1)j - 4i)$.

  (Solves $b = 6j - 5, 30j - 22, 72j - 51, 132j - 92, \ldots$.)