Find all prime numbers $ p $ for which $ 1 + p\cdot 2^{p} $ is a perfect square.

1. We need to find all prime numbers \( p \) such that \( 1 + p \cdot 2^p \) is a perfect square. Let's denote this perfect square by \( x^2 \). Therefore, we have:
   \[
   1 + p \cdot 2^p = x^2
   \]
   Rearranging, we get:
   \[
   x^2 - 1 = p \cdot 2^p
   \]
   This can be factored as:
   \[
   (x-1)(x+1) = p \cdot 2^p
   \]

2. Since \( x \) is an integer, both \( x-1 \) and \( x+1 \) are consecutive even numbers. Therefore, we can write:
   \[
   x-1 = 2a \quad \text{and} \quad x+1 = 2b
   \]
   where \( a \) and \( b \) are integers such that \( b = a + 1 \). Substituting these into the equation, we get:
   \[
   (2a)(2a+2) = p \cdot 2^p
   \]
   Simplifying, we have:
   \[
   4a(a+1) = p \cdot 2^p
   \]
   Dividing both sides by 4, we get:
   \[
   a(a+1) = \frac{p \cdot 2^p}{4}
   \]

3. Since \( p \) is a prime number, \( p \cdot 2^p \) must be divisible by 4. This implies \( p \) must be an odd prime (since \( 2^p \) is always even for \( p \geq 1 \)). Let's check small prime numbers to see if they satisfy the equation.

4. **Case \( p = 2 \):**
   \[
   1 + 2 \cdot 2^2 = 1 + 8 = 9 = 3^2
   \]
   So, \( p = 2 \) is a solution.

5. **Case \( p = 3 \):**
   \[
   1 + 3 \cdot 2^3 = 1 + 24 = 25 = 5^2
   \]
   So, \( p = 3 \) is a solution.

6. **Case \( p \geq 5 \):**
   Suppose \( p \geq 5 \) and \( 1 + p \cdot 2^p = x^2 \). Then:
   \[
   (x-1)(x+1) = p \cdot 2^p
   \]
   Since \( x-1 \) and \( x+1 \) are consecutive even numbers, we can write:
   \[
   \frac{x-1}{2} \cdot \frac{x+1}{2} = p \cdot 2^{p-2}
   \]
   Let \( y = \frac{x-1}{2} \) and \( z = \frac{x+1}{2} \). Then:
   \[
   y \cdot z = p \cdot 2^{p-2}
   \]
   Since \( y \) and \( z \) are consecutive integers, their greatest common divisor is 1. Therefore, one of them must be \( p \) and the other must be \( 2^{p-2} \). However, for \( p \geq 5 \), \( 2^{p-2} \) grows much faster than \( p \), making it impossible for \( y \) and \( z \) to be consecutive integers. Thus, there are no solutions for \( p \geq 5 \).

The final answer is \( \boxed{ p = 2 } \) and \( p = 3 \).