Question: First, how do you define "exhibit"? To you, would this mean you need to prove that your bijection is 1-1 and onto? If so, I think I've got the 1-1 part, but am struggling with showing it is onto. The examples of a linear function being shown as onto just seem too elementary, and I'm not sure they work when mapping naturals to integers.

I'm saying $f(n)=2n+13$. To show 1-1, I've let $f(x)=f(y)$, substituted through, and came out with $x=y$. I'm sure there needs to be more detail given, but for now, I'm just trying to remember how to show a function is onto. From what I've been able to find examples of, I basically solve my proposed function for $y$ getting $(y-13)/2$, then plug that in? Does that really show anything?! Does this work when going from $\mathbb{N} \rightarrow\mathbb{Z}$?

Answer: 
Your $f$ is a bijection $\Bbb N\to \{1,3,5,7,\dots\}$. This should be shifted with $12$ (if $13$ is included in the target set, and by $14$ if that starts with $15$), and you're there:
$$g(n):=f(n)+12=2n+13$$
will do it.
Suppose that $n\in \mathbb Z$ is odd and that $n>13$.

Then we can write $n=2z+1$ for some $z \in \mathbb Z$.

Then $n= 2z+1 =2z+1 +(13-13) = 2z-12 +13 = 2(z-6)+13$.

Hence $f(z-6) = n$.

Now we need to show that $z-6 \in \mathbb N$. But since $2z+1 > 13$ then $2z >12$ and $z>6$ and finally $z-6>0$.

Obviously all the greater than's can be greater than or equal to, according to whether or not $0\in \mathbb N$ for you.