Count the number of shortest paths to n
JavaScript (ES6),  111 ... 84  80 bytes
Returns true rather than \$1\$ for \$n=2\$.

f=(n,j)=&gt;(g=(i,x,e=1)=&gt;i?e&gt;n?g(i-1,x):g(i-1,x+e)+g(i,x,e*x):x==n)(j,2)||f(n,-~j)

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Commented
f = (                     // f is the main recursive function taking:
  n,                      //   n = input
  j                       //   j = maximum number of steps
) =&gt; (                    //
  g = (                   // g is another recursive function taking:
    i,                    //   i = number of remaining steps
    x,                    //   x = current sum
    e = 1                 //   e = current exponentiated part
  ) =&gt;                    //
    i ?                   // if there's still at least one step to go:
      e &gt; n ?             //   if e is greater than n:
                          //     add the result of a recursive call with:
        g(i - 1, x)       //       i - 1, x unchanged and e = 1
      :                   //   else:
                          //     add the sum of recursive calls with:
        g(i - 1, x + e) + //       i - 1, x + e and e = 1
        g(i, x, e * x)    //       i unchanged, x unchanged and e = e * x
    :                     // else:
      x == n              //   stop recursion; return 1 if x = n
)(j, 2)                   // initial call to g with i = j and x = 2
|| f(n, -~j)              // if it fails, try again with j + 1