Prove: $G$ is nilpotent iff $xy=yx \forall x,y\in G$ such that $(o(x),o(y))=1$
$G$ is finite. Is that plausible?
Attempt:
Suppose $G$ is nilpotnet. Then $G=P_1\times\ldots\times P_k$ where $P_1,...,P_k$ are the normal Sylow-$P$ subgroups of $G$. Let $x$ and $y$ be two elements of $G$ such that $(o(x),o(y))=1$.
If one of them is $1$ then we are done.
Otherwise, both $x,y \ne 1$. Since the orders have no common divisor that is not 1, $x$ and $y$ are composed by completely different $P_i$'s.
For example: If $x=p_1\cdot p_5\cdot p_{20}$ where $\{p_j\in P_j|j\in \{1,2,\ldots,20\}, p_j\ne 0\}$ then for $y$ of the aforementioned form $y=(y_1,y_2,\ldots,y_{20},\ldots,y_k)$, $y_1=y_2=\ldots=y_{20}=0$ necessarily.
Therefore $xy=x_1\cdot \ldots\cdot x_k \cdot y_1\ldots\cdot y_k$ where $x_i,y_i$ are either $0$ or in $P_i$ and if one is in $P_i$ the other one isn't. The reduced result of $x_1\cdot \ldots\cdot x_k \cdot y_1\ldots\cdot y_k$ will be a multiplication(or sum) of elements, each belong to a different $P_i$. Therefore they are commutative and we can get $xy=yx$.
Your idea is the correct one, but perhaps it'd be clearer as follows:
$$G=P_1\times\ldots\times P_k\;,\;\;P_i\;\;\text{the different Sylow subgroups of}\;\;G\;\implies$$
$$\forall\,g\in G\;\;\text{we can write uniquely}\;\;g=(x_1,...,x_n)\;,\;\;x_i\in P_i$$
and the operation is multiplication coordinatewise.
Now, if we have two elements with coprime orders
$$x=(x_1,..,x_k)\;,\;\;y=(y_1,..,y_k)\;,\;\;g.c.d(o(x)\,,\,o(y))=1$$
then it must be that for all $\;1\le i\le k\;$, either $\;x_i=1\;$ or $\;y_i=1\;$ (why?), and then the claim follows at once from the fact that
$$\forall\,i=1,2,...,k\;,\;\;\;[x_i,y_i]=1\implies [x,y]=1$$
and we're done.