Prove a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.

Here is a rigorous proof, but first we rewrite $f$:

$$f(m,n) = \frac12(m^2+2mn+n^2 + m + 3n) = \frac12((m+n)(m+n+1)+2n)$$

$\Large \textbf{Injectivity}$

Suppose we have $f(m,n) = f(a,b)$. Then $(m+n)(m+n+1)+2n = (a+b)(a+b+1)+2b$.

First, suppose $m+n\ne a+b$. WLOG suppose $m+n > a+b$. Then:

\begin{align}(m+n)(m+n+1)+2n &\ge (a+b+1)(a+b+2)\\&=(a+b)(a+b+1)+2a+2b+2 \\&>(a+b)(a+b+1)+2b \\&= (m+n)(m+n+1)+2n\end{align}

which is a contradiction. Hence $m+n=a+b$.

Using this fact we have $2n=2b$, and hence $(m,n) = (a,b)$.

$\Large \textbf{Surjectivity}$

Your table provides a great insight: $f(m,0)$ are precisely the triangular numbers, and $f(m-1, n+1) = 1+f(m,n)$ for $m > 0$.

We can prove this by: $$f(m,0) = \frac12(m^2+m) = T_m$$ \begin{align}f(m-1,n+1) &= \frac12((m-1+n+1)(m-1+n+1+1)+2(n+1))\\&=\frac12((m+n)(m+n+1)+2n)+1\\&=f(m,n)+1\end{align}

Now take any $x\in \mathbb N$. We can find a triangular number $T_k = \frac{k(k+1)}2$ such that $T_k \le x < T_{k+1}$.

Intuitively this $k$ would be $m+n$, and we need to shift over by $x-T_k$ numbers.

That is, notice that:

$$f(k-x+T_k, x-T_k) = \frac12((k)(k+1)+2(x-T_k))= T_k+x-T_k=x$$

This shows surjectivity.