In fact every number
z[m_] := 2^(2 + 4 m ) + 1
for m element Integers is divided by 5 without rest (proof by induction).
And some more solutions
2^# + 1 & /@ {2, 6, 10, 14}
This way you also get a negative answer:
In[83]:= Solve[Mod[5, 2^n + 1] == 0, n, Integers] Out[83]= {{n -> -2}, {n -> 2}}
