Let $p$ be an odd prime, and $a$, $b$ be two positive integers such that $2a>b$. If $4p^{2a}+4p^b+1$ is a perfect square, prove that $a=b$.
If $b<a$, then $(2p^a)^2<4p^{2a}+4p^b+1<(2p^a+1)^2$, which is contradictory. So $b\ge a$. If we write $4p^{2a}+4p^b+1=m^2$, then $(m+1)(m-1)=4p^b(p^{2a-b}+1)$. Since $(m+1,m-1)\le2$, then exactly one of $m+1$ and $m-1$ is divisible by $p^b$.
Then I don't know how to continue.
On
Your argument for the case $b < a$ is fine.
Next suppose $b > a$.
From $b > a$, we get $b\ge 2$, hence $$ p^b\ge 3^b=(2+1)^b > 2^b+b2^{b-1} > 2^b+2 $$ so $p^b-1 > 2^b+1$, an inequality to be used later.
As you observed, exactly one of $m+1,m-1$ is divisible by $p^b$.
Let $x$ denote the one of $m+1,m-1$ which is divisible by $p^b$, and let $y$ denote the other one.
Since $m$ is odd, $x,y$ are both even.
It follows that $x$ is divisible by $2p^b$.
Then $x\ge 2p^b$, hence from the equality $$ xy=(2p^b)\bigl(2(p^{2a-b}+1)\bigr)\qquad\qquad $$ we get \begin{align*} y\le\;&2(2^{2a-b}+1) \\[4pt] <\;& 2(2^b+1)\qquad\bigl(\text{since $b > a$}\bigr) \\[4pt] <\;& 2(p^b-1) \\[4pt] =\;& 2p^b-2 \\[4pt] \le\;& x-2 \\[4pt] \end{align*} so $y < x-2$, contradiction.
I continue with your solution. Since $4p^2a+4p^b+1$ is odd, then $m^2$ is odd implies m is odd. So $gcd(m+1,m-1) = 2$. Hence $4 \mid (m+1)(m-1)$ So we let $\frac{m+1}{2} = s$ and $\frac{m-1}{2} = s-1$ We have $s(s-1) = p^b(p^{2a - b} + 1)$ Since $(p^b, p^{2a - b} + 1) = 1$ and similar for $(s,s-1)$ = 1, then $$\\$$ Case 1: $p^b$ = $s-1$ and $p^{2a - b} + 1 = s$ $$\\$$Case 2: $p^b = s$ and $p^2a-b + 1 = s - 1$ $$\\$$ Case 1 implies a = b $$\\$$ For case 2 we have $p^b = p^{2a-b} + 2$, using the fact that $b\ge a$ we get $p^{2a-b}(p^{2b-2a} - 1) = 2$, $b = a$ leads contradiction so $b > a$, then $p^{2a-b} $ and $ p^{2b-2a} \ge 3$, contradiction because $p^{2a-b}(p^{2b-2a} - 1) = 2$.