Assume that $a \equiv b \text{ (mod m)}$, the following: $$a^n\equiv b^n \text{ (mod m)}$$ is true.
However, does $a^n\equiv b^n \text{ (mod m)}$ imply that $a \equiv b \text{ (mod m)}$ is true when n is odd?
If n is odd, it seems like this is justifiable because negative numbers do not become positive numbers if raised to an odd power.
Also would it be true for when n is even if say $a^n\equiv b^n \text{ (mod m)}$ imply $a \equiv ±b \text{ (mod m)}$?
The above example by Lord Shark the Unknown indicates that no complete answer can be found. But sometimes, we do have that kind of higher cancellation:
We have, by telescopic sum, $$ a^n - b^n = (a - b) \sum_{j=0}^{n-1} a^{n-1-j}b^j, $$ so that $a^n \equiv b^n \mod m$ implies $a \equiv b \mod m$ if $$ \sum_{j=0}^{n-1} a^{n-1-j}b^j $$ is not a zero divisor.
First, all you say about being positive or negative modulo $n$ makes really no sense, since $a\equiv a-m \pmod{m}$, so any positive number is equivalent to a negative number and viceversa.
However, there is a natural condition to put on $n$ related to $m$, if $m$ is prime: if $n$ and $m-1$ are coprime (so they share no comon divisor) , then $a^n\equiv b^n \pmod{m}$ implies $a \equiv b \pmod{m}$.
For $\gcd(n,\phi(m))=1$, then you will have $a^n\equiv b^n \implies a\equiv b \bmod m.$ (Since $\phi(m)$ is always even for $m>2$, $n$ will always be odd under this condition, but many odd numbers will not fulfill this).
When $\gcd(n,\phi(m))>1$, you can have two (or more) different values producing the same $n$th power - for example, $4^3\equiv 9^3\equiv 29 \bmod 35$.
