calculate the Integer solutions of an equation

Hint

The equation can be written as $$(2x+1)^2-(2y)^2=-247 \implies (2x+1-2y)(2x+1+2y)=-247.$$ Note that $247=13 \cdot 19$. So factor pairs are $(1,247), \, (13,19)$.

Multiply it by 4: $$4x^2+4x+248=4y^2$$ so $$(2x+1)^2+247 = 4y^2$$ and thus $$4y^2-(2x+1)^2=247$$