Concatenations of powers and their squares At time of writing, my reputation is \$16,256\$. As I noted in chat,

Oh cool my rep is the concatenation of two powers of 2: 16,256
Or even the concatenation of a power of 2 and its square, which is much more interesting

which then spawned a CMC about checking if a number has this property.

Given an integer \$n &gt; 0\$, considered a decimal integer, and a power \$r &gt; 1\$, return two distinct values which determine whether \$n\$ can be expressed as the concatenation of a power of \$r\$ and its square or not. For example, \$n = 16256\$ and \$r = 2\$ returns true (the concatenation of \$2^4\$ and \$(2^4)^2\$), while \$n = 39\$ and \$r = 2\$ does not. Note however that \$n = 39\$, \$r = 3\$ is true. The power of \$r\$ may be \$0\$, meaning that \$n = 11\$ is true for all \$r\$
The power of \$r\$ will always come &quot;before&quot; its square, so \$n = 62525, r = 5\$ is false.
You will never get an input \$n\$ where its validity depends on ignoring leading \$0\$s or not (for example \$101\$ is true for all \$r\$ if ignoring leading \$0\$s and false otherwise). However, you may still get inputs with the digit \$0\$ in (e.g. \$n = 1024, r = 2\$) where leading \$0\$s have no bearing on the validity of \$n\$ being such a concatenation.
Input and output may be in any accepted method and this is code-golf so the shortest code in bytes wins.
Test cases
    n    r    1
   39    3    1
  525    5    1
  864    8    1
16256    2    1
   11    r    1
  416    7    0
   39    2    0
   15    5    0
 1024    4    0
62525    5    0

Feel free to suggest more test cases.
APL (Dyalog Unicode), 21 bytes (SBCS)

Anonymous infix lambda, taking \$r\$ as left argument and \$n\$ as right argument. Requires ⎕IO←0 (zero-based indexing).
{⍵∊(⊢⍎⍤,⍥⍕¨×⍨)⍺*⍳⌊⍟⍵}

Try it online! (Dyalog Extended as polyfill for version 18.0)
{…} &quot;dfn&quot;, ⍺ is \$r\$ and ⍵ is \$n\$:
 ⍟⍵ natural log of \$n\$ (to avoid overflow)
 ⌊ round that down
 ⍳ ɩntegers zero through one less than that
 ⍺* raise \$r\$ to those powers
 (…) apply the following monadic function to that:
  ×⍨ multiply those with themselves (i.e. square them)
  ⊢…¨ for each unmodified argument and its corresponding square:
   ⍥⍕… stringify the argument and its square before
    ⍤, concatenating them, and then
     ⍎ evaluating the result
 ⍵∊ is the original argument a member of that?