How can I show that $4^{1536} - 9^{4824}$ can be divided by $35$ without remainder?

Notice that it's divisible iff $5$ and $7$ divide it. Look that

$$4^{1536}-9^{4824}\equiv_5 (-1)^{1536}-(-1)^{4824}= 1-1=0$$

then for the $7$, look that:

$$4^{1536}-9^{4824}\equiv_7 (16)^{768}-(2)^{4824}\equiv_7 (9)^{768}-(8)^{1608}\equiv (2)^{768}-(1)^{1608}$$

But $1^{1608}=1$, and $2^{768}=8^{256}$, so $$(2)^{768}-(1)^{1608}\equiv 8^{256}-1\equiv 1^{256}-1=0$$

For the case of division by $7$, I reduced the modules dividing by $2$ and trying to find a number congruent to $1$. If you use the little Fermat theorem, you could reduce the operations.

Euler's theorem implies, since $\varphi(35)=24$, that $$4^{24}\equiv 9^{24}\equiv 1\pmod {35}$$

Since $1536$ and $4824$ are multiples of $24$, the conclusion follows:

$$4^{1536}-9^{4824}=(4^{24})^{64}-(9^{24})^{201}\equiv1^{64}-1^{201}\equiv 0\pmod{35}$$

Since $35=7*5$ and $5,7$ are prime, it suffices to show that that is divisible by both $5$ and $7$. To do this, note that $4\equiv 9\equiv -1$ mod $5$, so $$ 4^{1536}-9^{4824}\equiv (-1)^{1536}-(-1)^{4824}\equiv 1-1\equiv 0\mod 5. $$ To show divisibility by $7$ requires a bit more work; try to show that successive powers of an element eventually cycle mod $7$, and then you will be able to get a nice form for them.