The answer is any $b$ such that $b+1$ is not a power of 2. In the forwards direction, we first prove more carefully the following claim.

Claim. If $b+1$ is a power of 2, then the only $n$ which is valid is $n=1$.

Proof. Assume $n > 1$ and let $p$ be the smallest prime dividing $n$. We cannot have $p = 2$, since then $4 \mid b^n + 1 \equiv 2 \pmod 4$. Thus,
$$b^{2n} \equiv 1 \pmod{p}$$
so the order of $b$ (mod $p$) divides $\gcd(2n, p-1) = 2$. Hence $p \mid b^2 - 1 = (b-1)(b+1)$.
But since $b+1$ was a power of 2, this forces $p \mid b-1$. Then $0 \equiv b^n + 1 \equiv 2 \pmod{p}$, contradiction. □

On the other hand, suppose that $b+1$ is not a power of 2 (and that $b > 2$). We will inductively construct an infinite sequence of distinct primes $p_0, p_1, \dots$, such that the following two properties hold for each $k \ge 0$:

* $p_0^2 \dots p_{k-1}^2 p_k \mid b^{p_0 \dots p_{k-1}} + 1$,
* and hence $p_0^2 \dots p_{k-1}^2 p_k^2 \mid b^{p_0 \dots p_{k-1} p_k} + 1$ by exponent lifting lemma.

This will solve the problem.

Initially, let $p_0$ be any odd prime dividing $b+1$. For the inductive step, we contend there exists an odd prime $q \notin \{p_0, \dots, p_k\}$ such that $q \mid b^{p_0 \dots p_k} + 1$. Indeed, this follows immediately by Zsigmondy theorem since $p_0 \dots p_k$ divides $b^{p_0 \dots p_{k-1}} + 1$. Since $(b^{p_0 \dots p_k})^q \equiv b^{p_0 \dots p_k} \pmod q$, it follows we can then take $p_{k+1} = q$. This finishes the induction.

To avoid the use of Zsigmondy, one can instead argue as follows: let $p = p_k$ for brevity, and let $c = b^{p_0 \dots p_{k-1}}$. Then $\frac{c^p+1}{c+1} = c^{p-1} - c^{p-2} + \dots + 1$ has GCD exactly $p$ with $c+1$. Moreover, this quotient is always odd. Thus as long as $c^p + 1 > p \cdot (c+1)$, there will be some new prime dividing $c^p + 1$ but not $c+1$. This is true unless $p=3$ and $c=2$, but we assumed $b > 2$ so this case does not appear.

Remark (On new primes). In going from $n^2 \mid b^n + 1$ to $(nq)^2 \mid b^{nq} + 1$, one does not necessarily need to pick a $q$ such that $q \nmid n$, as long as $\nu_q(n^2) < \nu_q(b^n + 1)$. In other words it suffices to just check that $\frac{b^{n+1}}{n^2}$ is not a power of 2 in this process.

However, this calculation is a little more involved with this approach. One proceeds by noting that $n$ is odd, hence $\nu_2(b^n + 1) = \nu_2(b+1)$, and thus $\frac{b^{n+1}}{n^2} = 2^{\nu_2(b+1)} \le b+1$, which is a little harder to bound than the analogous $c^p+1 > p \cdot (c+1)$ from the previous solution.