Let $m,n$ be natural numbers satisfying $m\leq 2n$. Is it true that $$2^{2n+2}+2^{m+2}+1$$ is a perfect square if and only if $m=n$?
What I have tried: Under the assumption $m<n$, I've tried to 'squeeze' the above number between two consecutive squares, implying it cannot be a perfect square. This works fine because (writing $P(m,n)=2^{2n+2}+2^{m+2}+1$), we have $(2\cdot2^n)^2<P(m,n)<(2\cdot2^n+1)^2$. But this method doesn't work when $\frac{m}{2}\leq n<m$, and I wonder if there is any pair $(m,n)$ with $m\ne n$ making $P(m,n)$ a perfect square.
Any advice is welcome.