Your conjecture is true! Here is one way to get it using some mod 4 generatingfunctionology. The answer got a bit long, so I divided it into two parts, as an attempt to improve readability.
Part1: Let's denote by $Y(q)$ your generating function $\sum_{i\geq 1}q^{y_i}$. We will prove that
$$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{16k})\pmod{2}.$$
Proof: One very classical theorem that dates back to Legendre states that the number of ways of writing $n\in \mathbb N$ as a sum of four triangular numbers is equal to $\sigma(2n+1)$, the sum of divisors of $2n+1$. Since the generating function of triangular numbers admits an infinite product expression from Jacobi's triple product formula, we can rewrite this theorem in the language of generating functions:
$$\sum_{n\geq 0}\sigma(2n+1)q^n=\prod_{k\geq 1}\frac{(1-q^{2k})^8}{(1-q^k)^4}.$$
I made use of this in an older answer, where we also pointed out the following simple corollary:
$$\sum_{n\geq 0}\sigma(4n+1)q^n\equiv\prod_{k\geq 1}\left(\frac{1-q^{4k}}{1-q^k}\right)^2=\prod_{k\geq 1}(1+q^{2k-1})^2(1+q^{2k})^4\pmod{4}.$$
Let us denote $$A(q)=\sum_{n\geq 0}\sigma((2n+1)^2)q^{\binom{n+1}{2}}.$$
The next lemma will allow us to split $\sum_{n\geq 0}\sigma(2n+1)q^n \pmod{4}$ into its even and odd parts. More precisely it will show
$$\sum_{n\geq 0}\sigma(2n+1)q^n\equiv A(q^2)+2qY(q^2) \pmod{4}$$
Lemma: The value of $\sigma(4n+1)$ is $1,3\pmod{4}$ iff 4n+1 is a perfect square, and it is $2\pmod{4}$ iff $n=2y_i+1$ for some $i$.
Proof of lemma: An odd number $m$ has a factorization $\prod p_i^{\alpha_i}q_j^{\beta_j}$ where $p_i$'s are primes $1\pmod 4$ and $q_j$'s are primes $3\pmod{4}$. The function $\sigma$ is multiplicative, so we just need to check the prime powers individually.
$$\sigma(p_i^{\alpha_i})=\frac{p_i^{\alpha_i+1}-1}{p_i-1}\equiv \alpha_i+1\pmod{4}$$
$$\sigma(q_j^{\beta_j})=\frac{q^{\beta_j+1}-1}{q_j-1}\equiv\left\{
\begin{array}{ll}
1\pmod{4} & \beta_j\text{ is even} \\
0\pmod{4} & \beta_j\text{ is odd} \\
\end{array}
\right.$$
Using the fact that $\sigma(m)=\prod \sigma(p_i^{\alpha_i})\sigma(q_j^{\beta_j})$ we see that $\sigma(m)$ is odd iff each $\alpha_i$ and $\beta_j$ are even, iff $m$ is a perfect square. We also see that $\sigma(m)=2\pmod{4}$ iff all $\beta_j$ are even, and exactly one $\alpha_i$ is $1\pmod{4}$, which is equivalent to saying that $m$ is one of the elements of your $x_i$ sequence. Now, $4n+1=x_i \implies n=2y_i+1$, so this concludes the proof of the lemma.
Some algebraic manipulations give us
$$\prod_{k\geq 1}(1+q^{2k-1})^2(1+q^{2k})^4=\prod_{k\geq 1}(1+q^{4k-2})(1+q^{2k})^4\prod_{k\geq 1}(1+2\frac{q^{2k-1}}{1+q^{4k-2}})$$
$$\equiv \prod_{k\geq 1}(1+q^{4k-2})(1+q^{2k})^4\left(1+2\sum_{k\geq 1} \frac{q^{2k-1}}{1+q^{4k-2}}\right)\pmod{4}.$$
Therefore we have an expression for $Y(q)\pmod{2}$
$$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\left(\sum_{k\geq 1} \frac{q^{k-1}}{1+q^{2k-1}}\right)\pmod{2}$$
$$\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\left(\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}\right)\pmod{2}.$$
Next we notice that the coefficient of $q^n$ in $\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}$ counts the number of integers $r\geq 0,k\geq 1$ such that $n=k-1+r(2k-1)$, which can be rewritten as $2n+1=(2k-1)(2r+1)$. The number of solutions to this equation is precisely the number of divisors of $2n+1$, and since the divisors of an odd number are odd, the number of divisors of $2n+1$ has the same parity as the sum of divisors of $2n+1$. Therefore $\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}$ is also equal to $\sum_{n\geq 0}\sigma(2n+1)q^n\pmod{2}$. Plugging this in our equation gives
$$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\frac{(1-q^{2k})^8}{(1-q^k)^4}\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{16k})\pmod{2}$$
and this finishes our proof. $\blacksquare$
Part2: Your conjecture would follow easily from the equality
$$\prod_{k\geq 1}\frac{1}{1-q^k}\equiv \prod_{i\geq 0}Y(q^{16^i})\pmod{2}.$$
Indeed the coefficient of $q^n$ on the left represents the parity of $p(n)$, the number of partitions of $n$, whereas the right hand side represents the number of ways of writing $n=z_0+16z_1+16^2z_2+\cdots$ where each $z_j$ is equal to some $y_i$.
In order to prove this we will use the mod 2 identity
$$\prod_{k\geq 1}(1+q^{2k-1})=\prod_{k\geq 1}\frac{1+q^k}{1+q^{2k}}=\prod_{k\geq 1}\frac{1+q^k}{(1+q^k)^2}=\prod_{k\geq 1}\frac{1}{1+q^k}\pmod{2}$$
which implies that
$$\prod_{k\geq 1}(1+q^{16k})=\prod_{k\geq 1}\frac{1}{1+q^{16(2k-1)}}\pmod{2}.$$
We can use this in our expression for $Y(q)$ to get
$$Y(q)=\prod_{k\geq 1}\frac{1+q^{2k-1}}{1+q^{16(2k-1)}}$$
which shows that the product $Y(q)Y(q^{16})Y(q^{16^2})\cdots$ telescopes to $\prod_{k\geq 1} (1+q^{2k-1})$ which is equal to the partition function $\pmod{2}$ thanks to the same identity above.
