We need to show that there is a bijection $h \colon \mathbb{Z}\_+ \times \mathbb{Z}\_+ \to \mathbb{Z}\_+$. For this purpose, Munkres define a subset $A$ of $\mathbb{Z}\_+ \times \mathbb{Z}\_+$ as follows: 
$$A \colon= \{ \ (x,y) \in \mathbb{Z}\_+ \times \mathbb{Z}\_+ \ \colon \ y \leq x \ \}.$$
Then he defines the maps $f \colon \mathbb{Z}\_+ \times \mathbb{Z}\_+ \to A$ and $g \colon A \to \mathbb{Z}\_+$ as follows: 
$$f(x,y) \colon= (x+y-1, y) \ \ \ \forall \ \ (x,y) \in \mathbb{Z}\_+ \times \mathbb{Z}\_+,$$
and 
$$g(x,y) \colon= \frac{(x-1)x}{2} + y \ \ \ \forall \ \ (x,y) \in A.$$
That the map $f$ is a bijection is clear to me. 

How to show (directly) that the map $g$ is also a bijection? 

It is clear intuitively. 

Once we show that $g$ is indeed a bijection, then the map $h$ could be defined as $$h \colon= g \circ f.$$

My effort: 

For all $(x,y) \in A$, we note that 
$$g(x,y) = y+ \sum\_{i=1}^{x-1} i.$$ 

What next?