I believe this is the entire solution set:

{10, 52, 64455910}

If ending in 00 were allowed, then every number of the form k * 10^(2k-1) for k= 1, 2, ..., 9 would be a solution:  10, 2000, 300000, 40000000 etc

Given the 00 restiction, here is a proof that no solution longer than 10 digits can exist.

9^9 = 387420489  which is 9 digits, so a 10 digit solution ending in 0 might be possible.

A 10 digit number consisting of all 9s yields 5*9^9 = 1937102445 which is 10 digits:  possible.

A 12 digit number consisting of all 9s yields 6*9^9 = 2324522934 which is 10 digits but reverse must be at least 11 digits:  impossible.

There can be no solution longer than 10 digits

I noticed that for a 10 digit solution, the 5 couplets would need to include at least one of the following,  in order to be at least 9 digits:  {79,89,99,98}  (although 7^9 is only 8 digits, 5 * 7^9 is 9 digits).  This sped up the the search of 10 digit numbers by about a factor of 5 or 6.

----------

ans = []

for d in range(2,11,2):

    for n in (range(10**(d-1), 10**d)):

        thesum = 0

        sn = str(n)

        if d == 10:

            prs = [sn[:2], sn[2:4], sn[4:6], sn[6:8], sn[8:10]]

            if not any(('79'in prs,'89'in prs,'99'in prs,'98'in prs)):

                continue

        target = int(sn[::-1])

        for ind in range(int(d/2)):

            thesum += int(sn[2*ind])  **   int(sn[2*ind+1])

        if thesum == target:

            print(n)

            ans.append(n)