Is there more than one?

let $k^2$ be the first of the ordered pairs,then the problem condition says that

$k^2+11111=x^2$ for some integer $x$ or equivalently $(x+k)(x-k)=11111$ .Now,since $11111=1*11111=41*271$ .these are the only possible ways you could factor $11111$ into two natural numbers.

we have $k=\frac{(271-41)}{2}$ or $k=\frac{(11111-1)}{2}$ .First one gives us solution: $(13225,24336)$ .The latter k is too big for $k^2$ to be a $5$ -digit number.

Hence,our answer is $\boxed{37561}$