df.iloc[i] returns the ith row of df. i does not refer to the index label, i is a 0-based index.

In contrast, the attribute index returns actual index labels, not numeric row-indices:

df.index[df['BoolCol'] == True].tolist()

or equivalently,

df.index[df['BoolCol']].tolist()

You can see the difference quite clearly by playing with a DataFrame with a non-default index that does not equal to the row's numerical position:

df = pd.DataFrame({'BoolCol': [True, False, False, True, True]},
       index=[10,20,30,40,50])

In [53]: df
Out[53]: 
   BoolCol
10    True
20   False
30   False
40    True
50    True

[5 rows x 1 columns]

In [54]: df.index[df['BoolCol']].tolist()
Out[54]: [10, 40, 50]

If you want to use the index,

In [56]: idx = df.index[df['BoolCol']]

In [57]: idx
Out[57]: Int64Index([10, 40, 50], dtype='int64')

then you can select the rows using loc instead of iloc:

In [58]: df.loc[idx]
Out[58]: 
   BoolCol
10    True
40    True
50    True

[3 rows x 1 columns]

Note that loc can also accept boolean arrays:

In [55]: df.loc[df['BoolCol']]
Out[55]: 
   BoolCol
10    True
40    True
50    True

[3 rows x 1 columns]

If you have a boolean array, mask, and need ordinal index values, you can compute them using np.flatnonzero:

In [110]: np.flatnonzero(df['BoolCol'])
Out[112]: array([0, 3, 4])

Use df.iloc to select rows by ordinal index:

In [113]: df.iloc[np.flatnonzero(df['BoolCol'])]
Out[113]: 
   BoolCol
10    True
40    True
50    True

Get the index for rows matching search term in all columns

search = 'security_id' 
df.loc[df.isin([search]).any(axis=1)].index.tolist()

Rows filtered for matching search term in all columns

search = 'search term' 
df.loc[df.isin([search]).any(axis=1)]

There might be more than one index map to your value, it make more sense to return a list:

In [48]: a
Out[48]: 
   c1  c2
0   0   1
1   2   3
2   4   5
3   6   7
4   8   9

In [49]: a.c1[a.c1 == 8].index.tolist()
Out[49]: [4]