<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] Here is a proof of finiteness, along the lines of Jordan's answer.  Assume without loss that $a>c$ and $d>b$.  Then
$$2^a-2^c=3^d-3^b,$$
or equivalently
$$\frac{1-3^{b-d}}{1-2^{c-a}}=\frac{2^a}{3^d}.$$
Since $a>d$, Matveev's explicit bound for linear forms in logarithms implies that
$$\Bigl\lvert\frac{2^{c-a}-3^{b-d}}{1-2^{c-a}}\Bigr\rvert=\lvert 2^a 3^{-d}-1\rvert \ge (ea)^{-R}$$
where 
$$
R:=e\cdot 2^{3.5}30^5\log 3.$$
Next, $2^c$ is the largest power of $2$ which divides $3^{d-b}-1$; if $2^i$ denotes the largest power of $2$ which divides $d-b$, then either $[i=0\,\text{ and }\,c=1]$ or $[i>0\,\text{ and }\,c=i+2]$.  Likewise, $b=0$ if and only if $a\not\equiv c\pmod{2}$; and if $a\equiv c\pmod{2}$ then $3^{b-1}$ is the largest power of $3$ which divides $a-c$.  So in any case we have $c\le 2+\log_2(d-b)$ and
$b\le 1+\log_3(a-c)$.  For any fixed value of $d$ there are only finitely many possibilities for $a$ (since $2^{a-1}\le 2^a-2^c<3^d$), and hence also for $b$ and $c$.  So assume that $d$ (and hence $a$) is sufficiently large; then $c\le 2+\log_2(d)<2+\log_2(a)$ and $b\le 1+\log_3(a)$ while $\log_3(2^{a-1})<d$, so $b\le\log_3(d)+r$ for some absolute constant $r$.  Now
$$\Bigl\lvert\frac{2^{c-a}-3^{b-d}}{1-2^{c-a}}\Bigr\rvert$$
is at most roughly $2^{-a}$, which is smaller than $(ea)^{-R}$ whenever $a$ is sufficiently large.
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