Finding the value of $ax^4+by^4$

Your method is not in vain. We have:

$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$

$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$

hence $x + y = 1, xy = -6$. Now:

$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$

$$691\times 1 = ax^4 + by^4 -6 \times 217$$

thus $ax^4 + by^4 = 1993$.