Suppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2-44x+12=p^n$. Find the largest value of $p$.
1. We start with the given equation:
   \[
   7x^2 - 44x + 12 = p^n
   \]
   We can factorize the left-hand side:
   \[
   7x^2 - 44x + 12 = (x-6)(7x-2)
   \]
   Therefore, we have:
   \[
   (x-6)(7x-2) = p^n
   \]

2. We need to find the largest prime number \( p \) such that the equation holds for some integer \( x \) and natural number \( n \).

3. Consider the case when \( x = 7 \):
   \[
   7(7)^2 - 44(7) + 12 = 7 \cdot 49 - 308 + 12 = 343 - 308 + 12 = 47
   \]
   Here, \( 47 \) is a prime number, and we can write:
   \[
   47 = 47^1
   \]
   Thus, \( p = 47 \) and \( n = 1 \).

4. Next, we need to check if there could be any larger prime \( p \) that satisfies the equation. For this, we analyze the factors of \( (x-6) \) and \( (7x-2) \).

5. Since \( (x-6)(7x-2) = p^n \), both \( x-6 \) and \( 7x-2 \) must be powers of \( p \). Let:
   \[
   x - 6 = p^a \quad \text{and} \quad 7x - 2 = p^b
   \]
   for some non-negative integers \( a \) and \( b \).

6. From \( x - 6 = p^a \), we get:
   \[
   x = p^a + 6
   \]
   Substituting \( x \) in \( 7x - 2 = p^b \):
   \[
   7(p^a + 6) - 2 = p^b
   \]
   Simplifying:
   \[
   7p^a + 42 - 2 = p^b
   \]
   \[
   7p^a + 40 = p^b
   \]

7. We need to check if there are any values of \( p \) larger than 47 that satisfy this equation. For \( p > 47 \), the left-hand side \( 7p^a + 40 \) grows much faster than the right-hand side \( p^b \), making it unlikely for the equation to hold for any \( a \) and \( b \).

8. Therefore, the largest prime \( p \) that satisfies the equation is \( p = 47 \).

The final answer is \( \boxed{47} \)