\[ (n-1)2^{n-1} + 5 = m^2 + 4m <=> (n-1) 2^{n-1} = (m-1)(m+5) \]
We shall for now skip the cases where n < 4 and m = 1, to note that (m-1) and (m-5) are congruent modular 2, but not congruent modular 4.
Therefore, either (m-1) or (m+5) can be written as: $ k . 2^{n-2} $ with k being an odd natural number.
If (m-1) = $ k . 2^{n-2} $, we replace m+5 by $ k . 2^{n-2} + 6 $
Replacing the 2 values to the above equation we get:
\[ (n-1) = k( 2^{n-3} +3) \]. Note that $ 2^b > b$ for every natural number b (including 0), so we come across a contradiction:
\[ n-1 >= 2^{n-3} + 3 > n-3 + 3 = n \]
If (m+5) = $ k . 2^{n-2} $, then with a similar reasoning, we get:
\[ n-1 = k( 2^{n-2} - 3) \].
If k is larger than 1, we get:
\[ n-1 >= 3. ( 2^{n-2} - 3) > 3 ( n - 2-3) = 3n - 15 \Rightarrow n < 7 \]You can solve for n, with 4 <= n < 7.
If k = 1, the equation becomes:
\[ n +2 = 2^{n-2} \]. Now we prove that: LHS < RHS for every n >=5.
(sorry for the bad formatting and clumsy language, I'm not a native)