Here are some ideas. If $n$ is a counterexample, then $m:=\sigma(n+2)-\pi(n)$ is a composite number such that $\sigma_2(m)-1$ is a square number ending in $1$. The goal would be to prove that there is no such $m$ (regardless of $n$).

Let us focus on the composite numbers $m$ such that $\sigma_2(m)-1$ is a square number. These numbers are listed at OEIS, and it seems that they are all even. Let us assume this, and let $k>0$ be the exponent of $2$ in $m$. I claim that $k$ is odd. To see this, observe that every divisor of $\sigma_2(2^k)$ is congruent to $1$ modulo $4$, because it is an odd number dividing $\sigma_2(m)$, which is a square number plus $1$. However, if $k$ is even, then $$\sigma_2(2^k)=\frac{4^{k+1}-1}{3}=\frac{(2^{k+1}+1)(2^{k+1}-1)}{3}$$ is divisible by $2^{k+1}-1$, which is congruent to $3$ modulo $4$. So $k$ is odd (under the standing assumption that $m$ is even). But this implies that $\sigma_2(2^k)$ is divisible by $5$, because in the previous display $4^{k+1}=16^{(k+1)/2}$ is congruent to $1$ modulo $5$. Hence $\sigma_2(m)$ is divisible by $5$, and therefore $\sigma_2(m)-1$ does not end in $1$.

To summarize, the OP's conjecture follows from the more natural conjecture that the linked OEIS sequence consists of even numbers.

I am now inclined to believe that the OP's conjecture is true, and I present some ideas to support this. If $n$ is a counterexample, then $m:=\sigma(n+2)-\pi(n)$ is a composite number such that $\sigma_2(m)-1$ is a square number ending in $1$. The goal would be to prove that there is no such $m$ (regardless of $n$).

Let us focus on the composite numbers $m$ such that $\sigma_2(m)-1$ is a square number. These numbers are listed at OEIS, and it seems that they are all even. Let us assume this, and let $k>0$ be the exponent of $2$ in $m$. I claim that $k$ is odd. To see this, observe that every divisor of $\sigma_2(2^k)$ is congruent to $1$ modulo $4$, because it is an odd number dividing $\sigma_2(m)$, which is a square number plus $1$. However, if $k$ is even, then $$\sigma_2(2^k)=\frac{4^{k+1}-1}{3}=\frac{(2^{k+1}+1)(2^{k+1}-1)}{3}$$ is divisible by $2^{k+1}-1$, which is congruent to $3$ modulo $4$. So $k$ is odd (under the standing assumption that $m$ is even). But this implies that $\sigma_2(2^k)$ is divisible by $5$, because in the previous display $4^{k+1}=16^{(k+1)/2}$ is congruent to $1$ modulo $5$. Hence $\sigma_2(m)$ is divisible by $5$, and therefore $\sigma_2(m)-1$ does not end in $1$.

To summarize, the OP's conjecture follows from the more natural conjecture that the linked OEIS sequence consists of even numbers.

Here are some ideas. If $n$ is a counterexample, then $m:=\sigma(n+2)-\pi(n)$ is a composite number such that $\sigma_2(m)-1$ is a square number ending in $1$. The goal would be to prove that there is no such $m$ (regardless of $n$).

Let us focus on the composite numbers $m$ such that $\sigma_2(m)-1$ is a square number. These numbers are listed at OEIS, and it seems that they are all even. Let us assume this, and let $k>0$ be the exponent of $2$ in $m$. I claim that $k$ is odd. To see this, observe that every divisor of $\sigma_2(2^k)$ is congruent to $1$ modulo $4$, because it is an odd number dividing $\sigma_2(m)$, which is a square number plus $1$. However, if $k$ is even, then $$\sigma_2(2^k)=\frac{4^{k+1}-1}{3}=\frac{(2^{k+1}+1)(2^{k+1}-1)}{3}$$ is divisible by $2^{k+1}-1$, which is congruent to $3$ modulo $4$. So $k$ is odd (under the standing assumption that $m$ is even). But this implies that $\sigma_2(2^k)$ is divisible by $5$, because in the previous display $4^{k+1}=16^{(k+1)/2}$ is congruent to $1$ modulo $5$. Hence $\sigma_2(m)$ is divisible by $5$, and therefore $\sigma_2(m)-1$ does not end in $1$.

Here are some ideas. If $n$ is a counterexample, then $m:=\sigma(n+2)-\pi(n)$ is a composite number such that $\sigma_2(m)-1$ is a square number ending in $1$. The goal would be to prove that there is no such $m$ (regardless of $n$).

Let us focus on the composite numbers $m$ such that $\sigma_2(m)-1$ is a square number. These numbers are listed at OEIS, and it seems that they are all even. Let us assume this, and let $k>0$ be the exponent of $2$ in $m$. I claim that $k$ is odd. To see this, observe that every divisor of $\sigma_2(2^k)$ is congruent to $1$ modulo $4$, because it is an odd number dividing $\sigma_2(m)$, which is a square number plus $1$. However, if $k$ is even, then $$\sigma_2(2^k)=\frac{4^{k+1}-1}{3}=\frac{(2^{k+1}+1)(2^{k+1}-1)}{3}$$ is divisible by $2^{k+1}-1$, which is congruent to $3$ modulo $4$. So $k$ is odd (under the standing assumption that $m$ is even). But this implies that $\sigma_2(2^k)$ is divisible by $5$, because in the previous display $4^{k+1}=16^{(k+1)/2}$ is congruent to $1$ modulo $5$. Hence $\sigma_2(m)$ is divisible by $5$, and therefore $\sigma_2(m)-1$ does not end in $1$.

To summarize, the OP's conjecture follows from the more natural conjecture that the linked OEIS sequence consists of even numbers.