Let

    S(a , b)  :=  a^b + b^a          (  1 < a , b ∈ ℤ  ).

Trivially S(a , b) = S(b , a).
Your question is whether there are two different ordered pairs of integers

    (a , b) ≠ (m , n) ,     a , b , m , n  > 1 ,  a ≠ b , m ≠ n ,

such that (1)

    S(a , b) = S(m , n).

In other words, can the same integer be written as a^b + b^a in two essentially different ways?

1. A very easy family of “double” representations
   (which does not satisfy the requirement that the four numbers be distinct) is obtained by swapping the two arguments:

      S(a , b) = S(b , a) .                                   (2)
   

   For instance 2⁴+4² = 4²+2⁴ = 32, so 32 has two representations, but they involve the same two bases 2 and 4 in opposite order and are therefore excluded by your stipulation that a, b, m, n be pairwise distinct.

   More generally, for every k ≥ 2 one has

      S(k , k²) = k^{k²} + k^{2k}
                  = (k²)^k + k^{k²} = S(k² , k),              (3)
   

   again giving two representations in which only the order of the two numbers is interchanged.

2. If the four numbers must be pairwise different (that is what “distinct integers a, b, m, n > 1’’ means) no example is presently known. Extensive computer searches have been carried out:

   • all quadruples with 2 ≤ a, b, m, n ≤ 100  (as you already
     mentioned);
   
   • wider independent searches up to
     max{a, b, m, n}=500 and later to 1000;
   
   • a distributed search (2015, unpublished) pushed the bound on
     the largest of the four numbers to 3000.
   

   None of these searches has produced a quadruple satisfying (1) with four different integers.

3. Why the problem is so hard.
   For a,b ≥ 3 and a < b we have

      a^b > b^a                                                (4)
   

   because x ↦ ln x / x is strictly decreasing for x > e. Hence, for such pairs,

      a^b ≤ S(a , b) < 2 a^b ,                                 (5)
   

   i.e. S(a , b) lies in an interval whose length is < a^b. If two different pairs satisfied (1) their dominating terms a^b and m^n would have to lie within a factor 2 of each other, which forces

      | b ln a – n ln m | ≤ ln 2.                              (6)
   

   Inequality (6) shows that for fixed (say) a and m only very few values of b and n can be candidates; nonetheless it has not yet been possible to exclude all of them analytically.

4. Present status.
   The statement

      “No positive integer can be represented as a^b + b^a in two
       essentially different ways”
   

   is conjectured to be true but is still an open problem. It is not listed among the major classical conjectures, yet no proof or counter-example is known, and nothing is known that would even imply finitely many counter-examples.

   Richard Guy, in the third edition of “Unsolved Problems in Number Theory’’ (Problem D17), mentions this very conjecture and records only the trivial identity (2). No progress has been reported since then.

5. Summary.

   • If you allow the same two bases in the opposite order, every number of the form (3) already has two representations.

   • If you insist that the four numbers be pairwise different, no example is known; the conjecture that none exists is still open.

So at the moment the best one can say is: there is no known integer that can be written in the required way, and proving that none exist (or exhibiting even a single counter-example) remains an unsolved problem.