Find all nonnegative integers $a$ and $b$ that satisfy the equation $$3 \cdot 2^a + 1 = b^2.$$
To find all nonnegative integers \(a\) and \(b\) that satisfy the equation 
\[ 3 \cdot 2^a + 1 = b^2, \]
we start by rewriting the equation in a factored form.

1. Rewrite the equation:
   \[ 3 \cdot 2^a + 1 = b^2 \]
   \[ 3 \cdot 2^a = b^2 - 1 \]
   \[ 3 \cdot 2^a = (b-1)(b+1) \]

2. Notice that \((b-1)\) and \((b+1)\) are two consecutive even numbers, and their product is \(3 \cdot 2^a\). Since one of these factors must be divisible by 4 (as they are consecutive even numbers), we analyze the possible values of \(b\).

3. Consider the possible values of \(b\):
   - If \(b = 2\):
     \[ (b-1)(b+1) = 1 \cdot 3 = 3 \]
     \[ 3 \cdot 2^a = 3 \]
     \[ 2^a = 1 \]
     \[ a = 0 \]
     Thus, \((a, b) = (0, 2)\).

   - If \(b = 5\):
     \[ (b-1)(b+1) = 4 \cdot 6 = 24 \]
     \[ 3 \cdot 2^a = 24 \]
     \[ 2^a = 8 \]
     \[ a = 3 \]
     Thus, \((a, b) = (3, 5)\).

   - If \(b = 7\):
     \[ (b-1)(b+1) = 6 \cdot 8 = 48 \]
     \[ 3 \cdot 2^a = 48 \]
     \[ 2^a = 16 \]
     \[ a = 4 \]
     Thus, \((a, b) = (4, 7)\).

4. Check if there are any other possible values of \(b\):
   - For \(b > 7\), the product \((b-1)(b+1)\) becomes too large to be equal to \(3 \cdot 2^a\) for nonnegative integer values of \(a\).

Therefore, the only solutions are:
\[ (a, b) = (0, 2), (3, 5), (4, 7) \]

The final answer is \( \boxed{ (a, b) = (0, 2), (3, 5), (4, 7) } \).