Is my understanding of Binary relations correct?

You are correct about $R$ being reflexive and not symmetric.

However, you have not proved that it is antisymmetric. You have only given a case that is antisymmetric, namely $(1,1)$, but you would need to show that it is true for all cases. You would need to prove the following statement:

For all $(a,b)\in \mathbb{Z}\times \mathbb{Z}$ the following holds: If $(a,b)\in R$ and $(b,a) \in R$, then $a=b$.

However, you will not be able to prove the statement because it is false. To demonstrate that it is false you need only one counter example. Can you see why any case of the form $(-a,a)$, where $a\neq 0$ makes it false?

Let $(a,b) \in \mathbb{Z}^2$.

If $(a,b) \in R$, then $a = m b$ for some $m \in \mathbb{Z}$.

If $(b,a) \in R$, then $b = n a$ for some $m \in \mathbb{Z}$.

So if $(a,b), (b, a) \in R$ then $$ a = m b = m (n a) = (m n) a \iff m n = 1 \iff m = n = 1 \vee m = n = -1 $$ The first case means $a = b$ and the second means $a = -b$.

The second case gives us counter examples. e.g. $(1, -1) \in R$ and $(-1, 1) \in R$, but $-1 \ne 1$.