Examples:
4 shares no factors with 7, and 4+7=11, which also shares no factors with 7
15 shares no factors with 16, and 15+16=31, which also shares no factors with 16
21 shares no factors with 17, and 21+17=38, which also shares no factors with 17
11 shares no factors with 13, and 11+13=24, which also shares no factors with 13
If $p$ is a number that divides both $n+m$ and $m$, then it also divides their difference, $(n+m)-m$, which is $n$. Since we assume that $m,n$ are coprime, then $p=1$.
First off, you should probably word your problem more rigorously, and say what you did or didn't try yourself. Also, use MathJax to format your math statements.
Anyway, the proof is as follows.
Suppose $n,m\in\mathbb{Z}$ such that $n$ and $m$ are coprime, that is to say, $(n, m)=1$. This notation means GCD of $n$ and $m$ is $1$, in case you didn't know, which is an equivalent statement to "sharing no factors".
Now, let $k$ be a shared factor of $n+m$ and $m$.
Then, $k|m$ and $k|n+m$.
So, there exists some $a,b\in\mathbb{Z}$, such that $ka = m$ and $kb = n+m$.
Substituting for $m$ in the second equation, we get $$kb = n + ka$$ $$ kb - ka = n $$ $$ k(b-a) = n $$
And since $n+m \gt m$, we know that $b \gt a$, so $b-a \gt 0$.
So, by definition, $k|n$.
Now, since $n$ and $m$ are coprime, and $k$ is a shared factor, it must be that $k|1$, which means that $k=1$.
Thus, any shared factor of $n+m$ and $m$ must necessarily be $1$, so $n+m$ and $m$ are coprime.
Also, note that $n+m$ would also be coprime with $n$ by the same logic.
