Find all natural values of $n$ and $m$, such that $(n -1)2^{n - 1} + 5 = m^2 + 4m$.
We are given the equation:

\[
(n - 1)2^{n - 1} + 5 = m^2 + 4m
\]

We need to find all natural values of \( n \) and \( m \) that satisfy this equation. Let's start by rewriting the equation in a more convenient form:

\[
(n - 1)2^{n - 1} + 5 = m^2 + 4m
\]

Subtracting 5 from both sides, we get:

\[
(n - 1)2^{n - 1} = m^2 + 4m - 5
\]

We can factor the right-hand side:

\[
(n - 1)2^{n - 1} = (m - 1)(m + 5)
\]

Now, we need to find natural numbers \( n \) and \( m \) such that the above equation holds. Let's analyze the factors on the right-hand side. Notice that \( (m - 1) \) and \( (m + 5) \) are both integers and their product must be equal to \( (n - 1)2^{n - 1} \).

### Case 1: \( m - 1 = k \cdot 2^{n-2} \)
Assume \( m - 1 = k \cdot 2^{n-2} \), where \( k \) is an odd natural number. Then:

\[
m + 5 = k \cdot 2^{n-2} + 6
\]

Substituting these into the equation, we get:

\[
(n - 1)2^{n - 1} = k \cdot 2^{n-2} \cdot (k \cdot 2^{n-2} + 6)
\]

Simplifying, we have:

\[
(n - 1) = k \left( 2^{n-3} + 3 \right)
\]

Since \( 2^{n-3} \) grows exponentially and \( n - 1 \) grows linearly, for \( n \geq 4 \), \( 2^{n-3} + 3 \) will always be greater than \( n - 1 \). This leads to a contradiction:

\[
n - 1 \geq 2^{n-3} + 3 > n - 3 + 3 = n
\]

Thus, this case does not provide any valid solutions.

### Case 2: \( m + 5 = k \cdot 2^{n-2} \)
Assume \( m + 5 = k \cdot 2^{n-2} \), where \( k \) is an odd natural number. Then:

\[
m - 1 = k \cdot 2^{n-2} - 6
\]

Substituting these into the equation, we get:

\[
(n - 1)2^{n - 1} = (k \cdot 2^{n-2} - 6) \cdot k \cdot 2^{n-2}
\]

Simplifying, we have:

\[
(n - 1) = k \left( 2^{n-2} - 3 \right)
\]

If \( k \geq 2 \), then:

\[
n - 1 \geq 2 \left( 2^{n-2} - 3 \right) > 2(n - 2 - 3) = 2n - 10
\]

This implies:

\[
n - 1 \geq 2n - 10 \Rightarrow n \leq 9
\]

We need to check values of \( n \) from 4 to 6.

### Checking specific values of \( n \):

1. **For \( n = 4 \)**:
   \[
   (4 - 1)2^{4 - 1} = 3 \cdot 8 = 24
   \]
   \[
   m^2 + 4m - 5 = 24 \Rightarrow m^2 + 4m - 29 = 0
   \]
   The discriminant is:
   \[
   \Delta = 4^2 - 4 \cdot 1 \cdot (-29) = 16 + 116 = 132
   \]
   Since the discriminant is not a perfect square, there are no integer solutions for \( m \).

2. **For \( n = 5 \)**:
   \[
   (5 - 1)2^{5 - 1} = 4 \cdot 16 = 64
   \]
   \[
   m^2 + 4m - 5 = 64 \Rightarrow m^2 + 4m - 69 = 0
   \]
   The discriminant is:
   \[
   \Delta = 4^2 - 4 \cdot 1 \cdot (-69) = 16 + 276 = 292
   \]
   Since the discriminant is not a perfect square, there are no integer solutions for \( m \).

3. **For \( n = 6 \)**:
   \[
   (6 - 1)2^{6 - 1} = 5 \cdot 32 = 160
   \]
   \[
   m^2 + 4m - 5 = 160 \Rightarrow m^2 + 4m - 165 = 0
   \]
   The discriminant is:
   \[
   \Delta = 4^2 - 4 \cdot 1 \cdot (-165) = 16 + 660 = 676
   \]
   Since the discriminant is a perfect square:
   \[
   \sqrt{676} = 26
   \]
   Solving the quadratic equation:
   \[
   m = \frac{-4 \pm 26}{2} \Rightarrow m = 11 \text{ or } m = -15
   \]
   Since \( m \) must be a natural number, \( m = 11 \).

Thus, the only solution is \( (n, m) = (6, 11) \).

The final answer is \( \boxed{ (n, m) = (6, 11) } \)