<s>[INST] Show that $(j,k)\mapsto 3k + j + (k+j)^2$ and $g$ are bijections from $\mathbb{Z}_{\ge 0}^2$ to $2\mathbb{Z}_{\ge 0}$. 
Show that $(j,k)\mapsto 3k + j + (k+j)^2$ and $(j,k)\mapsto 3j + k + (k+j)^2$ are two bijections from $\mathbb{Z}_{\ge 0}^2$ to $2\mathbb{Z}_{\ge 0}$.

Clearly both maps are well defined (consider congruences modulo $2$).
Let $g :\mathbb{Z}_{\ge 0}^2\to 2\mathbb{Z}_{\ge 0}$ be the described map. Note that $g$ is bijective if and only if $f :\mathbb{Z}_{\ge 0}^2 \to 2\mathbb{Z}_{\ge 0}, (j,k)\mapsto 3j + k + (k+j)^2$ is bijective. Indeed, if $n$ is a nonnegative even integer and $g^{-1}(n) = (j,k)$, then $f((k,j)) = n$. And if $f((j,k)) = f((s,t))$ then $g((k,j)) = g((t,s))\Rightarrow (k,j) = (t,s)$. The other direction is symmetric. So it suffices to show that $f$ is bijective.
Trying a few small values gives $(0,0)\mapsto 0, (0,1)\mapsto 2,(1,0)\mapsto 4,  (0,2)\mapsto 6, (1,1) \mapsto 8, (2,0)\mapsto  10.$ For fixed $x,$ there are $x+1$ pairs $(j,k)$ of nonnegative integers summing to $x$. For a given nonnegative even integer $y = 2n - 2, n \ge 1 1,$ find $x\ge -1$ so that $\frac{(x+1)(x+2)}2 < n \leq \frac{(x+2)(x+3)}2$. We claim that $f((i, x+1 - i)) = y,$ where $ i = n - \frac{(x+1)(x+2)}2 - 1 \ge 0$. Observe that $i \leq x+1$ so $(i, x+1-i)$ is indeed a pair of nonnegative integers. Also,
$\begin{align}f((i, x+1 - i)) &= 3i + (x+1-i)+(x+1)^2\\
&= 2i + (x+1)(x+2)\\
&= 2n -2.\end{align}$
To find x, first set $z = x+1$. Multiplying both sides of the inequality by $8$ gives $4z^2 + 4z < 8n \leq 4(z+1)(z+2) \Rightarrow (2z + 1)^2 < 8n + 1 \leq (2z + 3)^2\Rightarrow 2z + 1 < \sqrt{8n + 1} \leq 2z + 3\Rightarrow z <  \frac{\sqrt{8n+1}-1}2 \leq z + 1\Rightarrow z = \lceil\frac{\sqrt{8n+1}-3}2\rceil\Rightarrow x = \lceil\frac{\sqrt{8n+1}-5}2\rceil.$
Now we claim that all $(j,k)$ with $j+k\leq x$ get mapped to values that are at most $n$ and that if $n > \frac{(x+1)(x+2)}2,$ it appears in the list of pairs $(j,k)$ with $j+k = x+1$. In fact, if $n = \frac{(x+1)(x+2)}2$, then $n = f(x,0)$. Otherwise, if $n > \frac{(x+1)(x+2)}2$, then $n = f(i, x+1 - i)$, where $i = n - \frac{(x+1)(x+2)}2$.
From the above, we define the inverse of $y = 2n-2,n\ge 1$ to be $(i, x+1-i)$, where $i,x$ are given in terms of $n$ above.

Is there a simpler method?

 [/INST] Let $$f(j,k)=\frac{3k+j+(k+j)^2}2$$
Then $f(j-1,k+1)=f(j,k)+1$ and $f(k+1,0)=f(0,k)+1$, hence the assertion follows by induction.
For $f(0,0)=0$.
Assume $n=f(j,k)>0$.
If $j>0$, then $n+1=f(j-1,k+1)$.
Otherwise, $n=f(0,k)$, hence $n+1=f(k+1,0)$.
To prove uniqueness, note that if $j+k<j'+k'$, then
\begin{align}
f(j,k)&<f(0,j+k)\\
&<f(j+k+1,0)\\
&\leq f(j'+k',0)\\
&<f(j',k')
\end{align}
On the other hand, if $j+k=j'+k'$ and $j'<j$, then
\begin{align}
f(j,k)&<f(j',k+j-j')\\
&=f(j',k')
\end{align}
</s>