Hint $\ $ You have $\ a\,R\,b \iff a-b \in \{0\}\cup \{2^n\} =: S.\,$ Suppose more generally that $\,S\subset \Bbb Z\,$ is arbitrary set of integers containing $\,0.\,$ Then checking the equivalence relation properties
reflexive: $\quad\ \ \ a\, R\, a \iff 0\in S,\ $ which is true by hypothesis
symmetric: $\,\ (a\, R\, b\,\Rightarrow\, b\, R\, a)\iff (a\!-\!b\in S\,\Rightarrow\, b\!-a\!\in S)$
transitive: $\ \ \ (a\, R\, b,\ b\,R\,c\,\Rightarrow\, a\, R\, c)\iff (a\!-\!b,b\!-c\in S\,\Rightarrow\,a\!-\!c\in S)$
If so then $\,b=0\,$ in "symmetric" yields $\,a\in S\,\Rightarrow\, -a\in S,\,$ i.e. $\,S\,$ is closed under the operation of negation. Thus $\,a,c\in S\,\Rightarrow\,-c\in S\,$ so substituting $b,c \to 0,-c\,$ in "transitive" yields $\,a+c\in S,\,$ i.e. $\,S\,$ is closed under addition. Conversely, if $\,S\,$ is closed under addition and negation then then it satisfies the listed implications, so $\,R\,$ is an equivalence relation. If you know group theory then you will recognize this as the subgroup test, i.e. $\,R\,$ is an equivalence relation iff $\,S\,$ is a subgroup of the additive group of integers.
First, the relation is clearly reflexive since $a^2 - a^2 = 0 = a - a$ for any integer $a$.
Second, the equation $a^2 - b^2 = a - b$ is symmetric since switching $a$ and $b$ introduces negative signs on both sides. More precisely, if $a^2 - b^2 = a - b$ then $b^2 - a^2 = -(a^2 - b^2) = -(a - b)= b-a$.
Is it transitive? Suppose that $a^2 - b^2 = a - b$ and $b^2 - c^2 = b - c$. Then $a^2 - c^2 = a^2 - b^2 + b^2 - c^2 = (a^2 - b^2) + (b^2 - c^2) = (a-b) + (b - c) = a - c$, so indeed it is transitive.
To find the equivalence class of $5$ we need to find all integers $a$ such that $a^2 - 5^2 = a - 5$ which is equivalent to $0=a^2 - a - 20=(a + 4)(a-5)$, so we see that $a = -4$ or $a = 5$. Hence the equivalence class of $5$ consists of $\{ -4, 5\}$.
Best Answer
1) reflexive
$a=2^0\cdot a \Rightarrow aRa$
2) symmetric
if $a=2^kb$ then $b=2^{-k}a,$ where $k;-k \in \mathbb Z$
3) transitive
if $a=2^kb$ and $b=2^mc$ then $a=2^{k+m}c$ where $k;m, k+m \in \mathbb Z$