5     dim had(9)

  10     For k = 2 To 1549

  20       p4 = 4 ^ k

  30       p5 = 5 ^ k

  40       X1$ = Left(cutspc(Str(p4)), 1):v1=val(x1$)

  50       X2$ = Left(cutspc(Str(p5)), 1):v2=val(x2$)

  60       If X1$ = X2$ and had(v1)=0 Then

  70        :print k , Str(p4) , Str(p5) 

  72        :had(v1)=1

  80       :End If

  90     Next

  

finds only 4 and 2 as the possible values of x.

The first one for a value of 4 is at k=11 and the first one for a value of 2 is at k = 52. Other values of k show both x=4 and x=2. In the range tested, k= 2 through 1549 there were 74 occurrences of x=4 and 76 of x=2.  Beyond 1549, the fifth power caused overflow.

  

 11   4194304  48828125

 52   20282409603651670423947251286016  2220446049250313080847263336181640625