Solution:
We have to prove that $f: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ defined by
$$f(a, b) = \frac{1}{2}(a + b - 1)(a + b - 2) + a, \quad \forall a, b \in \mathbb{N}$$
is a bijection. (Note that the right side is a natural number.) To this end define
$$T(n) = \frac{n(n+1)}{2}, \quad n \in \mathbb{N} \cup \{0\}$$
An idea of the proof can be obtained by looking at the following table of values of $f(a, b)$ for some small values of $a, b$.

We observe that the $n$-th diagonal runs from $(1, n)$-th position to $(n, 1)$-th position and the entries are $n$ consecutive integers; the first entry in the $n$-th diagonal is one more than the last entry of the $(n-1)$-th diagonal. For example the first entry in 5-th diagonal is 11 which is one more than the last entry of 4-th diagonal which is 10. Observe that 5-th diagonal starts from 11 and ends with 15 which accounts for 5 consecutive natural numbers. Thus we see that $f(n-1, 1) + 1 = f(1, n)$. We also observe that the first $n$ diagonals exhaust all the natural numbers from 1 to $T(n)$. (Thus a kind of visual bijection is already there. We formally prove the property.)

We first observe that
$$f(a, b) - T(a + b - 2) = a > 0$$
and
$$T(a + b - 1) - f(a, b) = \frac{(a + b - 1)(a + b)}{2} - \frac{(a + b - 1)(a + b - 2)}{2} - a = b - 1 \geq 0$$
Thus we have
$$T(a + b - 2) < f(a, b) = \frac{(a + b - 1)(a + b - 2)}{2} + a \leq T(a + b - 1)$$
Suppose $f(a_1, b_1) = f(a_2, b_2)$. Then the previous observation shows that
$$\begin{aligned} & T(a_1 + b_1 - 2) < f(a_1, b_1) \leq T(a_1 + b_1 - 1) \\ & T(a_2 + b_2 - 2) < f(a_2, b_2) \leq T(a_2 + b_2 - 1) \end{aligned}$$
Since the sequence $\langle T(n) \rangle_{n=0}^{\infty}$ is strictly increasing, it follows that $a_1 + b_1 = a_2 + b_2$. But then the relation $f(a_1, b_1) = f(a_2, b_2)$ implies that $a_1 = a_2$ and $b_1 = b_2$. Hence $f$ is one-one.

Let $n$ be any natural number. Since the sequence $\langle T(n) \rangle_{n=0}^{\infty}$ is strictly increasing, we can find a natural number $k$ such that
$$T(k-1) < n \leq T(k)$$
Equivalently,
$$\frac{(k-1)k}{2} < n \leq \frac{k(k+1)}{2}$$
Now set $a = n - \frac{k(k-1)}{2}$ and $b = k - a + 1$. Observe that $a > 0$. Now (1) shows that
$$a = n - \frac{k(k-1)}{2} \leq \frac{k(k+1)}{2} - \frac{k(k-1)}{2} = k$$
Hence $b = k - a + 1 \geq 1$. Thus $a$ and $b$ are both positive integers and
$$f(a, b) = \frac{1}{2}(a + b - 1)(a + b - 2) + a = \frac{k(k-1)}{2} + a = n$$
This shows that every natural number is in the range of $f$. Thus $f$ is also onto. We conclude that $f$ is a bijection.