1. Squares of 2 different digits

Let b=a², b is consisted of 2 different digits.
First I describe the algorithm of counting the digits in b.

Let c$ be the string of b, that is c$=cutlspc(str(b)), and d is the length of b, d=len(c$).
Align z$(10) as the saving array of each digits.
Scaning c$ from rightmost digit and repeat the following procedure.

1. let f=1
2. first digit is stored in z$(1)
3. for second digit and after, compare to z$(1) through z$(f),
   • if there exist equal digit, do nothing.
   • otherwise, increse f and store the digit to z$(f).
4. when comparing to leftmost digit is complete, terminate the procedure.
   f indicates the amounts of different digits.

    b=a^2 : c$=cutlspc(str(b)) : d=len(c$)
    for i=d to 1 step -1 : f=0 : e$=mid(c$,i,1)
      if f=0 then inc f : z$(f)=e$ : goto *
      for j=1 to f
        if e$=z$(j) then cancel for : goto *
      next j
      inc f : z$(f)=e$
#     if f>2 then cancel for : goto ** : ' greater than 2
*   next i
    print a , b
**  return

Calling above routine from main loop of a, we can get the solutions of the arbitrary range.
The followings are the results of sporadic patterns.

a	b=a2
11 12 15 21 22	121 144 225 441 484
26 38 88 109 173	676 1444 7744 11881 29929
212 235 264 3114 81619	44944 55225 69696 9696996 6661661161

Only above 15 solutions are known.
I computed the range b ≤ 10⁴⁸, but couldn't find the next solution.

Above results are too depressing. So I extended the original problem as,

1. 3 different digits
2. perfect n-th power

For "3 different digits", change the "#" line as,

if f>3

and for "perfect n-th power", change the first line as,

b=a^n.