jmc

First, we note that if $n$ is a positive integer, then $f(n)$ is also a positive integer. We claim that $f(f(\cdots f(n)\cdots ))=1$ for some number of applications of $f$ only for $n=1,2,4,8,16,32,$ and $64.$ (In other words, $n$ must be a power of 2.)

Note that $f(1)=2,$ so $f(f(1))=f(2)=1.$ If $n>1$ is a power of 2, it is easy to see that repeated applications of $f$ on $n$ eventually reach 1.

Suppose $n$ is an odd positive integer, where $n>1.$ Write $n=2k+1,$ where $k$ is a positive integer. Since $n$ is odd, $$f(n)=n^2+1=(2k+1{)}^2+1=4k^2+4k+2=2(2k^2+2k+1).$$Since $2k^2+2k$ is always even, $2k^2+2k+1$ is always odd (and greater than 1), so $f(n)$ can never be a power of 2 when $n$ is odd and greater than 1.

Now, suppose $n$ is even. For example, if $n=2^3\cdot 11,$ then $$f(2^3\cdot 11)=f(2^2\cdot 11)=f(2\cdot 11)=f(11),$$which we know is not a power of 2.

More generally, suppose $n=2^e\cdot m,$ where $e$ is nonnegative and $m$ is odd. Then $$f(2^e\cdot m)=f(2^{e-1}\cdot m)=f(2^{e-2}\cdot m)=\cdots =f(m).$$If $m=1,$ then $n$ is a power of 2, and the sequence eventually reaches 1. Otherwise, $f(m)$ is not a power of 2. We also know that $f(m)$ is odd and greater than 1, $f(f(m))$ is not a power of 2 either, and so on. Thus, the sequence can never reach 1.

Therefore, $n$ must be one of the $\boxed{7}$ values 1, 2, 4, 8, 16, 32, or 64.