Irska

If $(ab{)}^2-4(a+b)=x^2$ with positive integers $a,b$ and an integer $x\ge 0$, we have $x<ab$. As $(ab{)}^2-(ab-1{)}^2=2ab-1$ is odd, we even have $x\le ab-2$. This implies $(ab{)}^2-4(a+b)\le (ab-2{)}^2=(ab{)}^2-4ab+4$, from which we obtain $$ab\le a+b+1.(5)$$ After swapping $a$ and $b$ if necessary, we may assume $a\le b$. If $a\ge 3$, we get $ab\ge 3b\ge a+b+b\ge a+b+1$ in contradiction to (5). Hence $a=2$ or $a=1$. If $a=1$, we have $b^2-4(b+1)=x^2$, which is equivalent to $(b-2-x)(b-2+x)=8$. Because $(b-2-x)+(b-2+x)=2b-4$ is even and $b-2-x\le b-2+x$, the only possibility is $b-2-x=2$ and $b-2+x=4$. This yields $(a,b)=(1,5)$ as the only possible solution with $1=a\le b$.

If $a=2$, we have $4b^2-4(b+2)=x^2$, equivalently $(2b-1-x)(2b-1+x)=9$. Here we have two possibilities. Either $2b-1-x=2b-1+x=3$ or $2b-1-x=1,2b-1+x=9$. In the first case we obtain $b=2$ and in the second $b=3$. So we have shown that $(a,b)=(2,2)$ and $(a,b)=(2,3)$ are the only possible solutions with $2=a\le b$.