I am given that $a,y$ positive integers and $$2^{1978}(1+2^{4a-2030})=y(y+1).$$ I want to maximize $a$.
I guessed that $a$ is maximized when $2^{1978}=y$ and hence $4a-2030=1978\implies a =1002$. However, there is the other case of when a factor of $1+2^{4a-2030}$ is "contained" in $2^{1978}$, that is: $(2^{1978}x, \frac{1+2^{4a-2030}}{x}) = (y,y+1) \text{ or }=(y+1,y)$. I have proven for the $(y,y+1)$ case that $2^{4a-30} \ge 2^{1978}x^2$ which imply that there may be infinite solutions and so there is no maximum.
If we multiply given equation $$2^{1978}(1+2^{4a-2030})=y(y+1)$$
with $4$ and then add $1$ we get $$\boxed{2^{1980}+2^{4a-50}+1=(2y+1)^2}$$
Now suppose there exists $a>1002$, then $$1980<2a-24$$ so $$\color{red}{2^{1980}+2^{4a-50}+1} <(2^{2a-25}+1)^2$$ and since $$2^{4a-50}<\color{red}{2^{1980}+2^{4a-50}+1}$$
so $$2^{4a-50}<\color{red}{(2y+1)^2}<(2^{2a-25}+1)^2$$ which implies $$2^{2a-25}<2y+1<2^{2a-25}+1$$
and thus no solution. So $a\leq 1002$ and thus $a_{\max} = 1002$ (which clearly works).