Why is every answer of $5^k - 2^k$ divisible by 3?

Yes, it does!

It's because in general you have the factorization:

$$ x^k-y^k = (x-y)(x^{k-1}+x^{k-2}y+\dots+y^{k-1}) $$

Substituting in $x=5$ and $y=2$ should show you why that works.

You say congruences are not familiar. Suppose you don't know the formulas in the other answers, but you do know the polynomial Division Algorithm. Dividing $\rm\:x^k\!-\!2^k$ by $\rm\:x\!-\!2\:$ yields

$\rm\qquad x^k-2^k =\ q(x)\, (x-2) + r\quad $ for an integer $\rm\:r\:$ and integer coefficient quotient $\rm\:q(x).$

Evaluating at $\rm\ x = 2\ $ shows $\rm\ r = 0.\ $ Evaluating at $\rm\: x = 5\:$ shows $\rm\,3\,$ divides $\rm\,5^k - 2^k$

Remark $\ $ This is a special case of the Factor Theorem, as are the other answers.

Recall that $$a^k - b^k = (a-b)(a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1}) \tag{$\star$}$$ One way to see this is, to notice that $$(a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1})$$ is the sum of first $k$ terms of a geometric progression with first term $a^{k-1}$ and common ratio $\dfrac{b}a$. We hence get that $$a^{k-1} +a^{k-2} b + \cdots + a b^{k-2} + b^{k-1} = a^{k-1} \left(\dfrac{1-(b/a)^k}{1-b/a}\right) = \dfrac{a^k-b^k}{a-b}$$ Rearranging above gives you $(\star)$.