This is (again) more a comment than an answer - motivated by René's question for more conceptional background
A couple of years ago I began to look at the full primefactorization of the cyclotomic polynomials $f_b(n) = b^n-1 $ by looking at $f(n)$ modulo the primes, creating a little "algebra" on it based on the theorems of Fermat ("little Fermat") and Euler ("Totient").
The following notations seem to be helpful for such an "algebra":
We're considering the canonical primefactorization of the expression
$$f_b(n) = p_1^{e_1} \cdot p_2^{e_2} \cdots p_m^{e_m} \tag 1$$
Looking at this for each primefactor $p_k$ separately ($f_b(n) \pmod {p_k}$) gives reason for two compact notations:
$[n:p]$ with the meaning $[n:p]=0$ if $p$ does not divide $n$ and $=1$ if it does divide $n$ (also known as "Iverson-brackets"; and no special definition for $n=0$ as long as not really needed)
$\{ n, p \} = e $ with the meaning of giving the exponent $e$, to which the primefactor $p$ occurs in $n$, so $ \{f_b(n),p_1 \} = e_1$ implies $f_b(n) = p_1^{e_1} \cdot x$ where $gcd(x,p)=1$ (in Pari/GP this is the function "valuation(n,p)")
The idea is to restate the defining equation (1) with the help of this notations/concepts. Of course, Fermat and Euler show us, that we have periodicity in the occurence of any primefactor, when we increase $n$ and that on special $n$ the primefactors $p_k$ occur even with higher exponent. To have expressive formulae for this too we introduce the formula for
- the smallest $n$ at which the primefactor $p$ occurs first in $f_b(n)$ (this is often written as $\text{ord}()$ denoting the "order of the multiplicative subgroup mod p" but to avoid possible conflicts with common terms we just call this $\lambda_b(p) $, so in $ f_b(\lambda_b(p)) $ the primefactor $p$ occurs first time when $n$ increases from $1$. Here and in the following we can remove the index-parameter $b$ at $f$ and at $\lambda$ for notational convenience, and even remove the argument with the parenthese at the $\lambda$-function when the referred prime $p$ is obvious from context. So we write $ [b^{\lambda(p)} -1 :p]=1$ (In Pari/GP it is $\lambda_b(p)=$
znorder(Mod(b,p)) )
We'll find, that sometimes in $f(\lambda(p))$ the primefactor $p$ occurs not only to the first, but by some higher power, so we introduce the function
- $\alpha_b(p)$ by the implicite definition $ \{ f_b(\lambda_b(p)),p \} = \alpha_b(p) $ or simplified $ \{ f(\lambda),p \} = \alpha $
For the odd primefactors $p$(the primefactor $p=2$ needs one extension) and of course when the base $b$ is coprime to the selected $p$, we can then state
$$ \{b^n-1 , p\} = [n:\lambda]\cdot (\alpha + \{n, p\}) \tag 2$$
For the primefactor $2$ and odd $b$ the $\lambda$-function is always $1$ . And because now always $[f(1):2]=1$ and also $[f(1)+2:2]=1$ the general expression (2) needs some refinement, but which I do not want show here - its indication may suffice for the following.
The question whether the same difference $a^x - b^y =d$ can also occur with $a^{x+v} - b^{y+w} =d$ can be rewritten as
$$ \begin{array}{rcl}
a^{x+v} - b^{y+w} &= &a^x -b^y \\
a^x(a^v-1) &=& b^y(b^w-1) \\
{a^v-1 \over b^y} &=& {b^w-1 \over a^x}
\end{array} \tag 3$$
we find that function $f_a(v)=a^v-1$ with divisibility condition by $b^y$ and similarly on the rhs.
The "conceptual" aspect is now, that on the lhs as well as on the rhs we have terms whose canonical primefactorizations are expressible by the above functions and notations
and must be equal except for the bases $a$ and $b$ which in the given examples are usually primenumbers (and thus primefactors of the other expression) themselves, for instance for the problem $3^{3+v}-5^{2+w} \overset{?}= 3^3-5^2=2$ and $a=3$ and $b=5$ here (and other examples as discussed in Will Jagy's answers) and possible values $\gt 0$ for $v$ and $w$ are sought.
Using the canonical primefactorizations we can write
$$ 3^v-1 = 2^{e_1} \cdot 3^0 \cdot 5^2 \cdot 7^{e_4} \cdots =\prod p_k^{e_k}\\
5^w-1 = 2^{h_1} \cdot 3^3 \cdot 5^0 \cdot 7^{h_4} \cdots = \prod q_i^{h_i} \\
$$
and for a solution all variable exponents must respectively be equal: $e_k=h_k$ to have equality in eq(3)
For searching a possible solution one can, a bit more than @WillJagy has done this, write down a sufficient list of primefactors and the compositions of $3^v-1$ and $5^w-1$ by that primefactors . With Pari/GP one can easily find
$$ \small \begin{array} {rl|rl}
\{3^v-1,2\} &= e_1 = 1+ [v:2] + \{v,2\} &
\{5^w-1,2\} &= h_1 = 2+ \{w,2\} \\
\{3^v-1,3\} &= e_2 = 0 &
\{5^w-1,3\} &= h_2 = [w:2](1+ \{w,3\}) \\
\{3^v-1,5\} &= e_3 = [v:4](1+ \{v,5\}) &
\{5^w-1,5\} & = h_3 = 0 \\
\{3^v-1,7\} &= e_4 = [v:6](1+ \{v,7\}) &
\{5^w-1,7\} &= h_4 = [w:6](1+ \{w,7\}) \\
\vdots
\end{array}$$
There are now two critical aspects in that list:
ansatz a) we must find some $v$ and $w$ such that all $e_k=h_k$ except $e_3=2$ and $h_2=3$ . But as we see, the $\lambda$-entries in the $[v:\lambda]$-terms have common divisors and so the inclusion of some primefactor $p_k$ means automatically the inclusion of another primefactor$ p_m$ due to the fact, that $\lambda(p_k)$ might contain $\lambda(p_m)$ as a divisor. And that inclusion would also imply the primefactor $q_m$ with the same exponent and thus the inclusion of other $q_n$ and so on. So this might run into an infinite progress and this would then give a contradiction to the assumption, that some pair of finite $(v,w)$ might allow a solution.
ansatz b) we must - in the logic of a)- find a pair $(v,w)$ which imply an inclusion of the bases as primefactors to an exponent which is higher than wanted, such that, for this example in the lhs the primefactor 5 is included to the power of 3 or in the rhs the primefactor 3 is included to the power of 4 or higher.
The case b) is the simpler one and can occur already when short lists of primefactors of $f_a(v)$ and $f_b(w)$ are checked after some $v$ and $w$ are recognized as mandatory to have equal primepowers at all.
The actual computation procedure is in principle the same as Will Jagy has done it, only that I provide an initial list of consecutive primes as possible primefactors of $f_a(v)$ and of $f_b(w)$ , keep their respective $\lambda_a(p_k), \lambda_b(q_k)$ and $\alpha_a(p_k),\alpha_b(q_k)$ .
The from the example inserted in (3)
$$ \begin{array}{rcl}
3^{3+v} - 5^{2+w} &= &3^3 -5^2 = 2 \\
3^3(3^v-1) &=& 5^2(5^w-1) \\
{3^v-1 \over 5^2} &=& {5^w-1 \over 3^3}
\end{array} $$
we have that $\{3^v-1,5\}=2$ is required, so by
$\{3^n - 1,5\} = [n:4](1+\{n,5\}) = 2 $ we find that $ [n:4]=1 $ and also $1+\{n,5\}=2$ and thus $n=4\cdot 5 = 20$ and thus the initial exponent $v_0$ must be set as $v_0=n=20$. Of course $v_0 = 20$ implies that other primefactors shall be included (and by hand we can simply create the list of that other primefactors by factorizing $factor(3^20-1)$ using Pari/GP) . What I've got including only the first 100 primes is
$$ \begin{array} {}
p_k & \lambda_3(p_k) & \alpha_3(p_k) & y'\\
2 & 1 & 1 & 1 \\
5 & 4 & 1 & 1 \\
11 & 5 & 2 & 2 \\
61 & 10 & 1 & 1
\end{array}$$
*(the column y' here means, that including the primefactor $p_k$ and using the thus required value of $v$ we get $5$ to the power of
y' in $f_3(v)$ )*
Similarly this can be done using $ \{5^w-1,3\} =3 $ following $ \{5^n-1,3\} = [n:2](1+\{n,3\}) = 3 \to n = 2 \cdot 3^2 $ and $w_0 = 18$. In the similar way as before we find, that other primefactors $q_k$ are now involved,see this:
$$\small \begin{array} {}
q_k & \lambda_5(q_k) & \alpha_5(q_k) & x' \\
2 & 1 & 2 & 2 \\
3 & 2 & 1 & 1 \\
7 & 6 & 1 & 2 \\
19 & 9 & 1 & 3 \\
31 & 3 & 1 & 2
\end{array} $$
Next, because all exponents of the involved primefactors $p_k$ and $q_k$ must be equal $e_k = h_k$ we build the common set $C$ of involved primefactors having the maximum exponent $c_k=max(e_k,h_k)$, excluding the primefactors which equal the mutual bases. That means, for instance, we have to increase $v_1$, such that $v_2=v_1 \cdot x$ and the prime $p = 31$ can occur in the list of $p_k$ with exponent $2$.
This is a very systematic job, given the above list of $\lambda$'s and $\alpha$'s and can be done using only a finite list of possible primefactors to include, say of length $100$.
This allows then a (relatively) simple algorithm which can be applied "blindly" to some problem.
1) Initialization: given the bases $a$ and $b$ select an upper bound maxk for primefactors in the primefactorization. Initialize the lists of $\lambda$ and $\alpha$ for $p_k$ and $q_k$ up to maxk primes with respect to base1 $b_1= 3$ and base $b_2 = 5$ and the required exponents $x=3$ and $y=2$. Compute the initial $v_1$ and $w_1$ from the condition, that $5^2$ shall be factor of $f_3(v)$ and $3^3$ shall be factor of $f_5(w)$
2.a) adaption: at iteration-step $i$ given $v_i$ produce the list of primefactors $p_k$ which would occur in $f_3(v_i)$ and given $w_i$ the list $q_k$ which would occur in $f_5(w_i)$ .
2.b) combination: create the combined list $C$ of all occuring primefactors with maximal occuring exponent and compute the required $v_{i+1}$ and $w_{i+1}$ which allow the occurence of all $C_k$ in $f_3(v_{i+1})$ and in $f_5(w_{i+1})$
Iterate the steps 2.a and 2.b until either in $f_3(v_i)$ are too many primefactors $p_3 =5$ or in $f_5(w_i)$ are too many primefactors $p_2=3$. If this does not occur in a meaningful number of iterations, increase the number maxk and start again or break with inconclusive result.
With two iterations of the steps 2.a and 2.b I get the following with some simple Pari/GP-procedures:
maxk=100;b1=3,b2=5;x=3;y=2
init (b1,b2, x,y, maxk)
\\ result: v=20 w=18 {f_3(v) -1, 5}= 2=y {f_5(w) -1, 3}= 3 =x
adapt
\\ primeslist p_k = [2, 5, 11, 61]
\\ primeslist q_k = [2, 3, 7, 19, 31]
\\result : v=360 w=1980 {f_3(v) -1, 5}= 2=y {f_5(w) -1, 3}= 3 =x
adapt
\\ primeslist p_k = [2, 5, 7, 11, 13, 19, 31, 37, 41, 61, 73, 181, 241, 271]
\\ primeslist q_k = [2, 3, 7, 11, 13, 19, 23, 31, 37, 41, 61, 67, 71, 89, 181, 199, 331, 397, 521]
\\result : v=720720 w=11880 {f_3(v) -1, 5}= 2=y {f_5(w) -1, 3}= 4 >x !!
\\ here we get now the contradiction because f_5(w) has too many factors 3
Other than in Will Jagy's code here is less "guess" - using a defined set of possible primefactors (the first
maxk primes) and only the iterated adapt-function seems to provide the contradiction-result without further manual intervention and/or guesses - so this is the reason, that I assume this a better solution reflecting a "conceptual" ansatz.
The Pari/GP-code is not difficult and I can append them on request.
(errors, typos shall be removed when I detect them)
[update]: the essay with more systematic explanations was updated
