Explicit injections are overrated. You actually have the knowledge to finish on your own, without writing this injection...
You already know that every function from $\mathbb Z^+$ to itself is a subset of the countable set $\mathbb Z^+\times\mathbb Z^+$. In particular this means that the collection of all functions from the positive integers to itself is a subset of $\mathcal P(\mathbb{Z^+\times Z^+})$.
Now you can use the fact that $\mathbb{|Z^+\times Z^+|=|Z^+|}$ and conclude that the power sets are equipotent, and therefore $\left|2^{\mathbb Z^+}\right|=\left|\mathbb{\left(Z^+\right)^{Z^+}}\right|$, as wanted.
If you still insist on finding this injection, let $\varphi\colon\mathbb{Z^+\times Z^+\to Z^+}$ be a bijection.
Now map $f\colon\mathbb Z^+\to\mathbb Z^+$ to the following function $G_f\colon\mathbb Z^+\to\{0,1\}$: $G_f(\varphi(a,b))=1\iff (a,b)\in f$
Show that $G_f$ is an injective function, it's not too hard but requires some element chasing and bookkeeping.