Bariz - Problem

We know that $f(0,y)=y+1$. We can then find formulae for $f(1,y)$, $f(2,y)$ and $f(3,y)$ in terms of $y$: \begin{align*} f(1,y)&=f(0,f(1,y-1))\\&=f(1,y-1)+1\\&=\cdots\\&=f(1,0)+y\\&=y+2\\ \end{align*}and similarly: $$f(2,y)=2y+3$$$$f(3,y)=2^{y+3}-3$$We know that $f(4,0)=13=2^4-3=2^{2^2}-3$. Assume, for induction, that $f(4,y)=\,^{y+3}2-3$ where $^ba$ represents $a^{a^{\iddots^a}}$ with a power tower of length $b$. We know that this holds true for $f(4,0)$ and that: $$f(4,y)=2^{f(4,y-1)-3}-3=2^{(\,^{y+2}2)-3+3}-3=\,^{y+3}2-3$$completing the induction. So, our answer is: $$\boxed{^{1984}2-3}$$ Edit: I was surprised this problem was not given a solution on this site despite being an IMO problem and several years old. So, I gave it one...