(I'll try make updates on this when I find something else)
hopefully I'm not wrong :\
Let the perfect power be $a^m$.
We can see that when $n=1$, $m=2$ will always work iff $b+4$ is a perfect square. ($b \geq 5$ is assumed throughout here).
When $n=m=2$, all $b \geq 5$ will work, as the result will always be $(b+2)^2$.
From now on, we will only be dealing with $n \geq 3$.
There are no perfect $m$th powers when $b=5$.
The following case is where $m$ is $\textbf{odd}$, and $\textbf{not a multiple of $5$.}$
We can rearrange the question to this form:
$$b^n + 4\frac{b^n-1}{b-1} = \frac{b^{n+1}+3b^n-4}{b-1} = a^m$$
Subbing $b=5$ yields $$\frac{8\cdot5^n-4}{4} = 2\cdot5^n-1 = a^m$$
Thus, $a$ must be odd.
$$2 \cdot 5^n = (a+1)(a^{m-1}-a^{m-2}+\cdots +a^2-a +1)$$
Thus, we know that $2 \mid a+1$. Since the rest are just powers of $5$, as well as $a+1 \leq a^2-a+1$ for $a \geq 2$, we have that
$$a+1 \mid 2(a^{m-1}-a^{m-2}+\cdots +a^2-a +1)$$
However, we also have that
$$a+1 \mid 2(a^{m-1}-a^{m-2}+\cdots +a^2-a -(m-1))$$
Thus, $$a+1 \mid 2\cdot m$$
Since $m$ is odd, $2$ cannot divide $m$. Since $m$ is also not a multiple of $5$, we must have that $$a+1 \mid 2$$
Since $(a+1)$ is also a factor of $2 \times 5^n$, we have that $a+1=2$ and so $a=1$. This gives us that $a \cdot 5^n = 2$ and so $n=0$, which is impossible.
The following case is where $m$ is $\textbf{odd}$, and $\textbf{is a multiple of $5$.}$
Now, if $m \equiv 0 \pmod 5$, and $m$ is odd, then let $m= 5^s \cdot M$, where $M$ is not divisible by $5$. Let ${a^5}^s = A$. Then, we have that $$a^{5m}= a^{5^s \cdot M} = ({a^{5^s}})^M = A^M$$
Again, we have that $$2 \cdot 5^n -1 = A^M$$
We can apply the solution above since $M$ is not a multiple of $5$. This gives us that $A+1=2$, and so $A=1$ and $n=0$, a contradiction.
If $m$ is a perfect power of $5$, then we will rearrange it slightly. Let $a^{5^{s-1}} = A$. Then, we have that $2 \cdot 5^n = a^5 + 1$. If we expand it, and use the above techinque, we get that $a+1 \mid 10$.
$a=9$ is the only option since $a$ has to be odd. $a=1$ leads to a contradiction. However, subbing it in reveals
$$2 \cdot 5^n = 9^5 - 1$$
Clearly, the RHS is divisible by $4$, but the LHS isn't, so this is a contradiction.
The following case is where $m$ is $\textbf{even}$.
We have that $m=2^r \cdot J$.
Like above, let $\alpha = {a^2}^r$. Then we have that
$$2 \cdot 5^n -1 = a^m = a^{2^r \cdot J} = (a^{2^r})^J = \alpha^J$$
Here, $J$ is odd, and so depending on whether $J$ is divisible by $5$, it can be dealt by the previous cases, unless $J=1$.
Assuming that $m \neq 2$, let $t = a^{2^{r-2}}$. We then have that $ 2\cdot 5^n - 1 = t^4$.
Multiplying both sides by $4$, we get
$$8 \cdot 5^n - 4 = 4t^4$$
Adding 16 to both sides yields
$$8 \cdot 5^n + 12 = 4t^4 + 2^4$$
Using Sophie Germain's identity https://en.wikipedia.org/wiki/Sophie_Germain%27s_identity :
$$8 \cdot 5^n + 12 = (2t^2+4t+4)(2t^2-4t+4)$$
Dividing by $4$ yields
$$2 \cdot 5^n + 3 = (t^2+2t+2)(t^2-2t+2)$$
Now, let's check all possible cases for $t$ modulo $5$.
$t \equiv 0$ yields RHS as $4$.
$t\equiv 1,2$ yields RHS as $0$ (First term $= 0$)
$t\equiv 3,4$ yields RHS as $0$ (Second term $= 0$)
Note that the LHS is $3$. Thus, there are no solutions.
Thus, we are left with only the $m=2$ case. The first lemma in this paper by John Cohn
https://www.cambridge.org/core/services/aop-cambridge-core/content/view/FC63F15647E84273C77B910D003C0192/S0017089500031207a.pdf/perfect_pell_powers.pdf
states that all solutions to the Diophantine equation $2z^k - 1= y^2$ where $k>2$ are $y=z=1$ and $y=239,z=13,k=4$. However, with our base being $5$, $z=5$ is not part of a solution and thus there are no squares.
There are no perfect $m$th powers when $b = 2k$, for $m \geq 3$.
Let $b=2 \cdot k$. We get that
$$(2k)^{n+1}+3(2k)^n-4=(2k-1)a^m$$
$$2^{n+1}k^{n+1}+3\cdot2^nk^n-4=(2k-1)a^m$$
Since $n \geq 3$, we can see that the LHS can be factored to form
$$4(2^{n-1}k^{n+1}+3\cdot2^{n-2}k^n-1)=(2k-1)a^m$$
Notice that the LHS is even, and $2k-1$ is odd, so $a^m$ must be even, and thus $a$ must be even. However, there are no more than 2 powers of $2$ on the LHS, so if $m \geq 3$, there will not be enough powers of $2$.
Thus, we are done.
There are no perfect $m$th powers when $v_2(b)=1,2$ unless $n=3$.
It suffices to show that there are no squares, from the result above.
Now, if $n \geq 4$ and $v_2(b)=1$:
Rewrite $b$ as $2j$.
We yield that $$2^{n+1}j^{n+1}+3\cdot2^nj^n-4=(2j-1)a^2$$
$$2^{n-1}j^{n+1}+3\cdot2^{n-2}j^n-1=(2j-1)A^2$$
where $a=2A$.
Now, note that $j$ is odd here. Thus, $2j -1 \equiv 1 \pmod 4$.
Now, taking $\pmod 4$ gives us that $-1 \equiv a^2 \pmod 4$ if $n-2 \geq 2$, which is the $n \geq 4$ requirement. However, this cannot happen.
This argument can be repeated similarly for $\pmod 8$, which leaves us with the case of $n=4$. Then, according to @Oscar_Lanzi's result, this contains no squares, and thus we are done.
