\(x^2+2x+1-\left(y^2+4y+4\right)=7\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x+y+3\right)\left(x-y-1\right)=7\)
Do x, y nguyên dương \(\Rightarrow x+y+3>3\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+3=7\\x-y-1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy pt có cặp nghiệm nguyên dương duy nhất \(\left(x;y\right)=\left(3;1\right)\)