Yesterday night I watched the video on youtube and tried my chance :) I was able to rule out some small number of cases.
If $b^{2a}-4a^b$ is a perfect square, then any $a>1$ must be even.
Proof. We use the already proven fact that $b^{2a}-4a^b$ is a perfect square iff there are positive integers $c,d$ satisfying the equations
\begin{align}
a^b &= cd \label{eq:cd} \tag{1} \\
b^a &= c+d \label{eq:c+d} \tag{2}.
\end{align}
Assume that $b^{2a}-4a^b$ is a perfect square and $a > 1$ is an odd number. Then, $c$ and $d$ are odd numbers by \eqref{eq:cd}. But then $b$ must be even by \eqref{eq:c+d}.
Let $b=2k$. Since $b^{2a}-4a^b$ is a perfect square, we have a Pythagorean triple of the form $p = (\star, 2a^k, (2k)^a)$. It is well-known that for any Pythagorean triple $(x,y,z)$ either all $x, y, z$ are even or $z$ is odd and $x$ and $y$ have different parity. Observe that simplifying the triple $p$ via dividing each term by 2 yields a new triple of the form $p' = (\star, a^k, 2^{a-1}k^a)$. No such Pythagorean triple can exist since $a^k$ is odd but $2^{a-1}k^a$ is not, a contradiction. $\tag*{\(\Box\)}$
OK, here is a follow-up result.
No solution exists for $b \ge 4a$.
Proof.
First, we need the following simple lemma:
If $b^{2a} - 4a^b \ge 0$ and $b > 2a$, then $a<k^\frac{2}{k-2}$ where $k = b/a$.
Proof.
By assumption there exists a real $k > 2$ such that $b=ka$. Then we have
\begin{align}
(ka)^{2a} &\ge 4a^{ka} \\
k^{2a}a^{2a} &> a^{ka} \\
k^{2a} &> a^{(k-2)a} \\
k^{2/(k-2)} &> a
\end{align}
$\tag*{\(\Box\)}$
It is obvious that all trivial solutions satisfy $b<4a$. It was proven that any other solution must satisfy $3 \le a < b$. Since we also proved that any $a>1$ must be even, we have $4 \le a < b$. Setting $b=4a$ in the lemma yields the bound $a<4$ for which there is no non-trivial solution. Note that $k^\frac{2}{k-2}$ is a decreasing function of $k$ for $k>2$. $\tag*{\(\Box\)}$
A simple brute-force search proves that any non-trivial solution must satisfy $1000 < a < b < 2.233a$.
There is no non-trivial solution if $b$ is an integer multiple of $a$. Specifically, the only possible case where $b=2a$ leads to $a=1, b=2$.
Proof. Let $b^{2a} - 4a^b = t^2$ for some positive integer $t$. Since $b=2a$, we have $(2a)^{2a} - 4a^{2a} = (2^{2a}-4)a^{2a} = t^2$. Observe that $a^{2a}$ is a square number. Thus, $2^{2a}-4$ must be a square as well. But $2^{2a}$ is also a square. There is a single pair of squares whose difference is $4$, which is $(0,4)$. Therefore, we have $a=1$. $\tag*{\(\Box\)}$
