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.