We find, empirically,
$$ 35 = 2^3+3^3 \qquad \qquad \text{and} \qquad (35 = 2^5+3^1)$$
The first option, using $2^m - 3^n=35$ from $|2^m-3^n|$ is easy. The second option, using $2^m-3^n=-35$ is not so easy, and I've not completed that part yet.
First option:
Equalling the first expression for 35 with our equation with unknowns we write $$ 2^m - 3^n = 2^3+3^3$$.
Reordering for equal bases on each side gives
$$2^m -2^3 = 3^n +3^3 $$
We see, that $m>3$ is required. So let $m=3+u$ then
$$2^3(2^u -1) = 3^n+3^3$$
But now, $\pmod {2^3}$, the residue of the lhs is zero but of the rhs one of $\{4,6\}$. So no solutions is possible.
Second option:
We use the second equation:
$$ 3^n-2^m = 35 = 2^5+3^1 $$
Then we can reorder
$$ 3^n - 3^1 = 2^5+2^m $$
By the minimal magnitude of the rhs it follows that $n \ge 4$ and from this follows that $m \ge 5$ because $3^4-3 = 78 > 2^5+2^5$. Then we can factor
$$ 3(3^u-1) = 2^5(2^v+1) \qquad \qquad \text{ setting $u=n-1$ and $v=m-5$ }$$
Now to have $(3^u-1)$ containing $32=2^5$ as factor we need that $2^3$ is factor of $u$ but not $2^4$ - otherwise we had too many powers of $2$ here. So we need to have
$$ 3(3^{8+16u'}-1) = 2^5(2^v+1) $$
Next, to have in the rhs $3$ to the first power, but not to the second, we must have the exponent being odd and in particular of the form $6v' \pm 1$ so we must have exponents fulfilling the equation
$$ 3(3^{8+16u'}-1) = 2^5(2^{ 6v' \pm 1}+1) $$
But having such a regular exponent on the lhs includes also the primefactor $5$ which occurs always when the exponent is divisible by $4$, and this is always the case in this configuration.
On the other hand, for the same primefactor $5$ to appear as well in the rhs, we must have an even exponent and namely of the form $2^{2+4v"}$
So, for to have in the rhs exactly one factor $3$ the exponent must be odd, but to have the rhs also to contain the factor $5$ the exponent must be even - contradiction.
Comment: well, this version might look no so nice/smooth as the other one which uses only a single modulus $85$ for both options. But we have here smaller moduli ($8$ in the first option and $64$ and $9$ and $5$ in the second option) and we have some path which leads
systematically to the contradiction (if it exists) by some sort of
"iterative mutually adapting the exponents" and checking the occuring primefactors beginning with the smallest occuring ones.