After nesting the $2^x(2y+1)-1$ pairing function into its linear variable some finite number of times, the final linear component will have to be 0 eventually for any finite starting integer. I think this arbitrary-dimensional tupling function based on that pairing function might be an example of what you are talking about:
Tuple(x1,x2,...,xn)=(2^(n-1))*((2^sum(xj,j=1:n))-sum(2^(-k+sum(xj,j=k+1:n)),k=1:n-1))-1
$$\operatorname{Tuple} (x_1, x_2, \ldots, x_n) = 2^{n-1} \times
\left(
\left(2^{\sum_{j=1}^n x_j}\right)
-\sum_{k=1}^{n-1} 2^{-k+\sum_{j=k+1}^n x_j}
\right)-1$$
as
Tuple(x1,x2,...,xn) = Tuple(0,0,0,...,x1,x2,...,xn) = Tuple(x1,x2,...,xn)
$$\operatorname{Tuple} (x_1, x_2, \ldots, x_n) = \operatorname{Tuple} (0, 0, 0, \ldots, x_1, x_2, \ldots, x_n) = \operatorname{Tuple} (x_1, x_2, \ldots, x_n)$$
I hope this is helpful and correct, and I’m curious to know applications for this type of thing. I found your question researching this, and a similar function which can be nested into its linear variable in (almost) the same way:
P(x,y)=(y+1)*Fib(x+3)+floor(y/phi)*Fib(x+2)-2
$$P(x,y) = (y+1) \operatorname{Fib} (x+3) +\lfloor y/\phi \rfloor \operatorname{Fib} (x+2) -2$$
where Fib(n) is the nth Fibonacci number, and phi is the golden ratio (1+sqrt(5))/2.
It is harder to write down the expression of the nested version, but it is easy to program, and easy to find a quick algorithm to unpair a given integer repeatedly until its linear component is 0.
This function is tricky to turn continuous while maintaining its values on the integers, whereas the first function is already set up to be continuous.
I’d love to know a use for some of this!