Conditionally Remove Brackets

Another solution using xparse.

\documentclass{article}

\usepackage{xparse}

\NewDocumentCommand{ \Af }{ o }{%
    A\IfValueT{#1}{_{(#1)}}%
}
\NewDocumentCommand{ \Bf }{ d() }{%
    A\IfValueT{#1}{_{(#1)}}%
}

\begin{document}
no argument: $\Af$

with argument: $\Af[x]$

no argument: $\Bf$

with argument: $\Bf(x)$
\end{document}

xparse uses a different (an in my eyes easier and more flexible) syntax to define a command. The second argument of \NewDocumentCommand takes a list defining the arguments, e.g. o for an optional an m for a mandatory argument. With \If(No)Value(TF) you can test whether the argument has a value. (TF) can be T or F to test only one case or TF to differentiate between both the latter case has there arguments, e.g. {#1}{true code}{false code}. With d as argument specifier (see \Bf definition) you can even define the delimiting symbols, () in this case, so the code equals the output even more.

As daleif said you can add a \smash to the subscript to prevent it form affecting the line spacing.

\NewDocumentCommand{ \Afs }{ o }{%
    A\IfValueT{#1}{_{\smash{(#1)}}}%
}

without \smash (exaggerating example using \sum and \left…\right to get flexible parens).

with \smash

Since you want an optional argument I think you should use the presence of the argument (rather than it being empty) as the test so

\makeatletter

\newcommand\Af{\@ifnextchar[\@Af\@@Af}

\def\@@Af{\mathbf{A}}
\def\@Af[#1]{\@@Af_{(#1)}}

\makeatother

...

\Af ... \Af[x] ...

I usually use etoolbox

Something like (note your example does not use an optional argument)

 \newcommand\Af[1][]{\mathbf{A}\ifblank{#1}{}{_{(#1))}}}

ought to do the trick (untested)

Usage:

$\Af$  will just give $\mathbf{A}$

and

$\Af[n]$ will give $\mathbf{A}_{(n)}$

BTW: One can do even cooler things with the xparse package such as instead of $\Af[n]$ we can use $\Af_{n}$ to get the same result (see the argument type e in the xparse manual), aka we pick up the n from the subscript and wrap () around it.