If $x=y$, then $x^2=4x+3x \implies x=7 \implies (x,y)=(7,7)$ If $x>y$, then $x^2=4y+3\gcd(x,y) \leq 7y<7x \implies x<7$. Since $y\geq 1$ and $\gcd(x,y)\geq 1$, $x^2\geq 7\implies x=3$, $4$, $5$ or $6$. Checking all cases, we have $(x,y)=(3,2)$, $(5,5)$, $(5,6)$. If $x<y$, then take $x=dx_1$ and $y=dy_1$, where $d=\gcd(x,y)$ and $\gcd(x_1,y_1)=1$ [\hide]