We prove that for any odd integer $m > 0$, there are infinitely many positive integers $n$, such that $b_{n+1} - b_n = m$. For sake of this reason we will use the following lemma.

Lemma 1. Let $a$ be a positive integer which is not a perfect square. There exist infinitely many primes $p \equiv 3 \pmod{4}$ such that $a$ is quadratic non-residue modulo them.

Proof. Let $p_1p_2\cdots p_s$ be the square free part of integer $a$, where $p_i$'s are distinct primes. Now choose the quadratic non-residue $r_1$ modulo $p_1$ and set $r_2, \dots, r_s$ as quadratic residues modulo $p_2, \dots, p_s$, respectively. Now by Chinese Remainder Theorem and Dirichlet's Theorem one can find infinitely many primes $p$ satisfying
$$p \equiv r_i \pmod{p_i} \quad \text{for } i = 1, 2, \dots, s$$
Note that if one of the primes named $p_1$ is equal to $2$, then omit the first congruence and just add the criterion $p \equiv 3 \pmod{8}$ and we are done. $\square$

Now set $m = 2M + 1$ and consider the sequence $k^2 + M^2, k^2 + M^2 + 1, \dots, k^2 + M^2 + 2M + 1$. In this sequence the first and the last term are represented as sum of two squares. We will prove that there are infinitely many integers $k$ such that only these two terms could be representable as sum of two squares. According to the lemma, there exist primes $p_1, p_2, \dots, p_{2M} \equiv 3 \pmod{4}$ such that for each $j$, $M^2 + j$ is quadratic non-residue modulo $p_j$. Now we establish the following lemma.

Lemma 2. Let $r$ be a quadratic residue modulo prime $P$ and $(x, P) = 1$. Then there is integer $x$ such that $0 < x < 2P$, $x^2 \equiv r \pmod{P}$ and $x^2 \not\equiv r \pmod{P^2}$.

Proof. Let us note that $(x + P)^2 - x^2 = P^2 + 2xP \not\equiv 0 \pmod{P^2}$. Then one of $x$ and $x + P$ satisfies the desired conditions. $\square$

Now by this lemma there exist positive integers $n_1, n_2, \dots, n_{2M}$ such that $0 < n_j < 2p_j$ ($1 \le j \le 2M$) and we have
$$n_j^2 \equiv -(M^2 + j) \pmod{p_j}, \quad x^2 \not\equiv -(M^2 + j) \pmod{p_j^2}$$
Next, let $k \equiv n_j \pmod{p_j^2}$ for $1 \le j \le 2M$. Then, the numbers $k^2 + M^2 + j$ are all divisible by $p_j$ and not $p_j^2$. Now, since $p_j \equiv 3 \pmod{4}$, $k^2 + M^2 + j$ cannot be written as sum of two squares. So we are done.