jmc

It may not be obvious how to proceed with this problem, but a little experimentation might lead you to determine the possible values of $m.$

Since $0<m^2<22,$ we can see that $m$ must be one of $1,$ $2,$ $3,$ or $4.$ So let's use these as our cases.

Case 1: When $m=1$, we must have $n<22-1=21.$ Thus there are $20$ possible choices for $n$ when $m=1.$ Case 2: When $m=2,$ we must have $n<22-4=18.$ Thus there are $17$ possible choices for $n$ when $m=2.$ Case 3: When $m=3,$ we must have $n<22-9=13.$ Thus there are $12$ possible choices for $n$ when $m=3.$ Case 4: When $m=4,$ we must have $n<22-16=6.$ Thus there are $5$ possible choices for $n$ when $m=4.$

So to get the total number of pairs of positive integers satisfying the inequality, we add up all of our possible cases, and see that there are $20+17+12+5=\boxed{54}$ possible pairs.