It's interesting!!

Write the equation as $y^2 = \frac{4428 + x^2}{1+x}$ . Then, the first condition $x$ must satisfy is $1 + x | 4428 + x^2 \implies 1 + x | 4428 - x \implies 1 + x | 4429$ . Fortunately for us, $4429 = 43 \times 103$ so we only have to check four positive divisors (negative divisors would give negative values for $x$ ) to see which of them yields a valid for value for $y$ (a perfect square).

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Optional:

Notice that the equation $y^2 + xy^2 - x^2 \equiv 4428 \mod 6$ has unique solution $x=y=0$ so we may say $x = 6x_1, y=6y_1$ and write the equation as $6^2 y_1^2 + 6^3 x_1 y_1^2 - 6^2x_1^2 = 4428 \implies y_1^2 + 6x_1 y_1^2 - x_1^2 = 123 \implies y_1^2 = \frac{123 + x_1^2}{1 + 6x_1}$ . So for a given value of $x$ , we may test that $y_1^2$ is a perfect square instead of $y^2$ . If we were doing this problem by hand, this would be useful as the numbers would be smaller.

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As $1+x | 4429$ , $1+x = 1, 43, 103, 4429 \implies x = 0,42,102,4428 \implies x_1 = 0, 7, 17, 738$ . We may ignore the value $x_1 = 0$ as we want positive solutions, so $x_1 = 7, 17, 730$ . Now, $\frac{123 + 7^2}{1 + 6 \times 7} = 4 = 2^2, \frac{123 + 17^2}{1 + 6 \times 17} = 4 = 2^2, \frac{123 + 738^2}{1 + 6 \times 738} = 123 = 2 \times 3 \times 41$ . We notice that all but the last one yield a perfect square, so the valid values of $x_1$ are $7, 17$ so the valid values of $x$ are $42, 102$ .