Solve the following system of equations for real $x,y$ and $z$:
\begin{eqnarray*}
x &=& \sqrt{2y+3}\\
y &=& \sqrt{2z+3}\\
z &=& \sqrt{2x+3}.
\end{eqnarray*}
1. Given the system of equations:
   \[
   \begin{cases}
   x = \sqrt{2y + 3} \\
   y = \sqrt{2z + 3} \\
   z = \sqrt{2x + 3}
   \end{cases}
   \]
   We start by squaring each equation to eliminate the square roots:
   \[
   \begin{cases}
   x^2 = 2y + 3 \\
   y^2 = 2z + 3 \\
   z^2 = 2x + 3
   \end{cases}
   \]

2. Rearrange each equation to isolate the terms involving \(x\), \(y\), and \(z\):
   \[
   \begin{cases}
   x^2 - 2y = 3 \\
   y^2 - 2z = 3 \\
   z^2 - 2x = 3
   \end{cases}
   \]

3. Notice that if \(x = y = z\), then substituting \(x\) for \(y\) and \(z\) in the original equations should satisfy all three equations. Let's test \(x = y = z\):
   \[
   x = \sqrt{2x + 3}
   \]
   Squaring both sides:
   \[
   x^2 = 2x + 3
   \]
   Rearrange to form a quadratic equation:
   \[
   x^2 - 2x - 3 = 0
   \]
   Factor the quadratic equation:
   \[
   (x - 3)(x + 1) = 0
   \]
   Thus, the solutions are:
   \[
   x = 3 \quad \text{or} \quad x = -1
   \]

4. Since \(x\), \(y\), and \(z\) must be real and non-negative (as they are under square roots in the original equations), we discard \(x = -1\). Therefore, we have:
   \[
   x = y = z = 3
   \]

5. Verify the solution \(x = y = z = 3\) by substituting back into the original equations:
   \[
   \begin{cases}
   3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3 \\
   3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3 \\
   3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3
   \end{cases}
   \]
   The solution \(x = y = z = 3\) satisfies all three equations.

6. To check for any other possible solutions, consider the case where none of \(x\), \(y\), or \(z\) is equal to 3. Multiply the three equations:
   \[
   (x^2 - 2y)(y^2 - 2z)(z^2 - 2x) = 27
   \]
   This simplifies to:
   \[
   (x^2 - 2y)(y^2 - 2z)(z^2 - 2x) = 27
   \]
   Since \(x, y, z \geq 0\), the left-hand side must be non-negative. However, the product of three non-negative terms equaling 27 implies that each term must be at least 3, which is not possible unless \(x = y = z = 3\).

Therefore, the only solution is \(x = y = z = 3\).

The final answer is \(\boxed{x = y = z = 3}\).