To solve the equation \( x^2 + 2x + 3 = (2x + 1)\sqrt{x^2 + 3} \), we can follow these steps:

1. **Square both sides** to eliminate the square root:
\[
(x^2 + 2x + 3)^2 = (2x + 1)^2 (x^2 + 3)
\]

2. **Expand both sides**:
- Left side:
\[
(x^2 + 2x + 3)^2 = x^4 + 4x^3 + 10x^2 + 12x + 9
\]
- Right side:
\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]
Then, multiplying by \( (x^2 + 3) \):
\[
(4x^2 + 4x + 1)(x^2 + 3) = 4x^4 + 12x^2 + 4x^3 + 12x + x^2 + 3
\]
\[
= 4x^4 + 4x^3 + 13x^2 + 12x + 3
\]

3. **Set the equation**:
\[
x^4 + 4x^3 + 10x^2 + 12x + 9 = 4x^4 + 4x^3 + 13x^2 + 12x + 3
\]

4. **Rearranging terms** gives:
\[
0 = 3x^4 + 0x^3 + 3x^2 - 6
\]
\[
0 = 3x^4 + 3x^2 - 6
\]
Dividing everything by 3:
\[
0 = x^4 + x^2 - 2
\]

5. **Let** \( y = x^2 \):
\[
0 = y^2 + y - 2
\]

6. **Factoring** gives:
\[
(y - 1)(y + 2) = 0
\]

This yields:
1. \( y - 1 = 0 \) \( \Rightarrow y = 1 \) \( \Rightarrow x^2 = 1 \) \( \Rightarrow x = 1 \) or \( x = -1 \)
2. \( y + 2 = 0 \) \( \Rightarrow y = -2 \) (not valid for real numbers)

7. **Final solutions**:
\[
x = 1 \quad \text{or} \quad x = -1
\]

8. **Check** these solutions in the original equation:
- For \( x = 1 \):
\[
LHS: 1^2 + 2(1) + 3 = 6, \quad RHS: (2(1) + 1)\sqrt{1^2 + 3} = 3 \cdot 2 = 6. \quad \text{holds true.}
\]
- For \( x = -1 \):
\[
LHS: (-1)^2 + 2(-1) + 3 = 2, \quad RHS: (2(-1) + 1)\sqrt{(-1)^2 + 3} = -1 \cdot 2 = -2 \quad \text{not valid.}
\]

Therefore, the only valid solution is:
\[
\boxed{1}
\]