Example 15 (2003 Beijing Competition Question) Given that $x, y$ are positive real numbers, and satisfy $x y+x+y=71, x^{2} y+x y^{2}=880$, then $x^{2}+y^{2}=$ $\qquad$ .
Solve: Fill in 146. Reason: Both the conditions and the desired result involve the values of two letters $x$ and $y$, with no difference. The goal is to reduce the difference in form and degree between the conditions and the desired result by exploring the common form they share. Since $x y + x + y = (x y) + (x + y), x^2 y + x y^2 = (x y)(x + y), x^2 + y^2 = (x + y)^2 - 2(x y)$, we can derive the following results:

Let $a = x + y, b = x y, x y + x + y = a + b = 71, x^2 y + x y^2 = x y (x + y) = a b = 880$.
Thus, $a$ and $b$ are the roots of the equation $t^2 - 71 t + 880 = 0$.
Solving for $a$ and $b$, we get $a = x + y$ and $b = x y$ as 16 and 55, respectively.
If $x + y = 55$ and $x y = 16$, there are clearly no positive integer solutions. Therefore, we have $x + y = 16$ and $x y = 55$.
Hence, $x^2 + y^2 = (x + y)^2 - 2 x y = 146$.