Find all real pairs $(x,y)$ that solve the system of equations \begin{align} x^5 &= 21x^3+y^3
\\ y^5 &= x^3+21y^3. \end{align}
We are given the system of equations:
\[
\begin{align}
x^5 &= 21x^3 + y^3, \\
y^5 &= x^3 + 21y^3.
\end{align}
\]

1. **Case 1: \( y = 0 \)**

   If \( y = 0 \), substituting into the first equation gives:
   \[
   x^5 = 21x^3.
   \]
   Factoring out \( x^3 \) from the equation, we get:
   \[
   x^3(x^2 - 21) = 0.
   \]
   This implies:
   \[
   x^3 = 0 \quad \text{or} \quad x^2 = 21.
   \]
   Therefore, \( x = 0 \) or \( x = \pm \sqrt{21} \). However, substituting \( x = \pm \sqrt{21} \) into the second equation with \( y = 0 \) does not satisfy the equation. Thus, the only solution in this case is:
   \[
   (x, y) = (0, 0).
   \]

2. **Case 2: \( y \neq 0 \)**

   Let \( a = \frac{x}{y} \). Then \( x = ay \). Substituting \( x = ay \) into the system of equations, we get:
   \[
   (ay)^5 = 21(ay)^3 + y^3,
   \]
   \[
   y^5 = (ay)^3 + 21y^3.
   \]
   Simplifying these equations, we have:
   \[
   a^5 y^5 = 21a^3 y^3 + y^3,
   \]
   \[
   y^5 = a^3 y^3 + 21 y^3.
   \]
   Dividing both equations by \( y^3 \) (since \( y \neq 0 \)), we get:
   \[
   a^5 y^2 = 21a^3 + 1,
   \]
   \[
   y^2 = a^3 + 21.
   \]
   Dividing the first equation by the second, we obtain:
   \[
   \frac{a^5 y^2}{y^2} = \frac{21a^3 + 1}{a^3 + 21},
   \]
   which simplifies to:
   \[
   a^5 = \frac{21a^3 + 1}{a^3 + 21}.
   \]
   Multiplying both sides by \( a^3 + 21 \), we get:
   \[
   a^5 (a^3 + 21) = 21a^3 + 1.
   \]
   Expanding and simplifying, we have:
   \[
   a^8 + 21a^5 = 21a^3 + 1.
   \]
   Rearranging terms, we get:
   \[
   a^8 + 21a^5 - 21a^3 - 1 = 0.
   \]
   Factoring the polynomial, we get:
   \[
   (a-1)(a+1)(a^2 + 3a + 1)(a^4 - 3a^3 + 9a^2 - 3a + 1) = 0.
   \]
   The last term \( a^4 - 3a^3 + 9a^2 - 3a + 1 \) is always positive, so it does not contribute any real roots. Therefore, the possible values of \( a \) are:
   \[
   a = 1, \quad a = -1, \quad a = \frac{-3 + \sqrt{5}}{2}, \quad a = \frac{-3 - \sqrt{5}}{2}.
   \]

   - For \( a = 1 \):
     \[
     x = y \quad \text{and} \quad y^2 = 22 \implies y = \pm \sqrt{22}.
     \]
     Thus, the solutions are:
     \[
     (x, y) = (\sqrt{22}, \sqrt{22}), (-\sqrt{22}, -\sqrt{22}).
     \]

   - For \( a = -1 \):
     \[
     x = -y \quad \text{and} \quad y^2 = 20 \implies y = \pm \sqrt{20}.
     \]
     Thus, the solutions are:
     \[
     (x, y) = (\sqrt{20}, -\sqrt{20}), (-\sqrt{20}, \sqrt{20}).
     \]

   - For \( a = \frac{-3 + \sqrt{5}}{2} \):
     \[
     y^2 = 2(3 + \sqrt{5}) \implies y = \pm 2\sqrt{3 + \sqrt{5}}.
     \]
     Thus, the solutions are:
     \[
     (x, y) = \left( \frac{-3 + \sqrt{5}}{2} y, \pm 2\sqrt{3 + \sqrt{5}} \right).
     \]

   - For \( a = \frac{-3 - \sqrt{5}}{2} \):
     \[
     y^2 = 2(3 - \sqrt{5}) \implies y = \pm 2\sqrt{3 - \sqrt{5}}.
     \]
     Thus, the solutions are:
     \[
     (x, y) = \left( \frac{-3 - \sqrt{5}}{2} y, \pm 2\sqrt{3 - \sqrt{5}} \right).
     \]

The final answer is \( \boxed{ (0,0), (\sqrt{22}, \sqrt{22}), (-\sqrt{22}, -\sqrt{22}), (\sqrt{20}, -\sqrt{20}), (-\sqrt{20}, \sqrt{20}), \left( \frac{-3 + \sqrt{5}}{2} y, \pm 2\sqrt{3 + \sqrt{5}} \right), \left( \frac{-3 - \sqrt{5}}{2} y, \pm 2\sqrt{3 - \sqrt{5}} \right) } \).