<s>[INST] How many solutions to $2^a + 3^b = 2^c + 3^d$? A few weeks ago, I asked on math.stackexchange.com how many quadruples of non-negative integers $(a,b,c,d)$  satisfy the following equation:
$$2^a + 3^b = 2^c + 3^d \quad (a \neq c)$$
I found 5 quadruples:  $5 = 2^2 + 3^0 = 2^1 + 3^1$, $11 = 2^3 + 3^1 = 2^1 + 3^2$, $17 = 2^4 + 3^0 = 2^3 + 3^2$, $35 = 2^5 + 3^1 = 2^3 + 3^3$, $259 = 2^8 + 3^1 = 2^4 + 3^5$
I didn't get an answer, but only a link to an OEIS sequence (no more quadruples below $10^{4000}$), so I'm asking the same question here. 

Is there a way to prove that they are [not] infinite?

And, more generally, are there known tuples for which the following equation:
$$p_{i_1}^{a_1} + p_{i_2}^{a_2} + ... + p_{i_n}^{a_n}=p_{i_1}^{b_1} + p_{i_2}^{b_2} + ... + p_{i_n}^{b_n}$$
holds for infinitely many  (or holds only for finitely many) $a_i,b_i$?
 [/INST] The answer is yes, there are finitely many for positive $a, b, c, d$, and you've found all of them.
See Theorem 7 of R. Scott, R. Styer, On the generalized Pillai equation $\pm a^x \pm b^y = c$, Journal of Number Theory, 118 (2006), 236–265.  I quote:

Theorem 7. Let $a$ be prime, $a>b$, $b = 2$ or $b = 3$, $a$ not a large base-$b$ Wieferich prime, $1 \le x_1 \le x_2$, $1 \le y_1 \le y_2$, and $(x_1, y_1) \neq (x_2, y_2)$.  If there is a solution $(a, x_1, y_1, x_2, y_2)$ to the equation $$\left|a^{x_1} - b^{y_1}\right| = \left|a^{x_2} - b^{y_2}\right|$$ then it is one of
$$\begin{align*}
3-2&=3^2-2^3,\\ 
2^3-3&=2^5-3^3,\\
2^4-3&=2^8-3^5,\\
\cdots&=\cdots
\end{align*}$$
where the omitted cases are irrelevant.  The three listed equations are your decompositions of $\fbox{11}$, $\fbox{35}$, and $\fbox{259}$.

Now we show that the answer is yes for non-negative $a, b, c, d$, and you've also found all of them.
The only solutions to your equation not covered by the above result correspond to solutions of $$1 - 2^x + 2^y = 3^z, $$ where we may assume $0 < x < y$.
If $z$ is odd, then the RHS is 3 modulo 4 and so we must have $x = 1$.  Hence $3^z - 2^y = 3 - 2$ and so by the above we have the only solution as $(x, y, z) = (1,2,1)$.  This is your decomposition of $\fbox{5}$.
If $z$ is even, then the RHS is a perfect square.  Working modulo 3 we see that we must have $x$ odd and $y$ even.  Write $z' = z/2, y' = y/2, x' = x-1$, and note that $x' > 0$ is even.
If $x' < y'$ then we have $(2^{y'}-1)^2 = 1 - 2^{y'} + 2^y < 1 - 2^x + 2^y < 2^y = (2^{y'})^2$, and so the LHS cannot be a perfect square.  Hence $x' \ge y'$.
If we have $x' = y'$, then we have $3^{z'} = 2^{y'} - 1$; this gives the solution $(x,y,z) = (3,4,2)$, corresponding to your decomposition of $\fbox{17}$.

We claim that there are no other solutions.
Any other solution must have $x' > y'$.  We rearrange and write $$\left(2^{x'} - 1 + 3^{z'}\right)\left(2^{x'} - 1 - 3^{z'}\right) = 2^{2x'} - 2^{2y'}.$$  Note that we must have $2^{x'}-1 > 3^{z'}$ so that the signs on each side match.
The RHS is divisible by $2^{2y'}$.  As $x < y$ we have $2y' \ge x' + 2$.
We have $\gcd(2^{x'} - 1 + 3^{z'}, 2^{x'} - 1 - 3^{z'}) = \gcd(2^{x'} - 1 - 3^{z'},2\cdot3^{z'})$, which is divisible by 2 but not by 4.  Hence one of $2^{x'} - 1 + 3^{z'}$ and $2^{x'} - 1 - 3^{z'}$ is divisible by $2^{2y' - 1} \ge 2^{x' + 1}$.  But $$2^{x' + 1} = 2^{x'} + 2^{x'} > 2^{x'} - 1 + 3^{z'} > 2^{x'} - 1 - 3^{z'},$$ and so neither can be divisible by $2^{2y' - 1}$.  Hence there are no more solutions.
</s>