I claim the answer is in fact $q-1=4p$. Note that $2^{4p}\equiv 1\pmod{q}$, and $i\mid 4p$ (a well-known fact). Clearly $p>2$ and $q\equiv 5\pmod{8}$. As $i\mid 4p$, we are left with $i\in\{1,2,4,p,2p,4p\}$. The cases $i=1,2,4$ are impossible since $p\ge 3$ so $q\ge 13$. Now if $i=p$ then $2^p\equiv 1\pmod{q}$, so $x^2\equiv 2\pmod{q}$ for $x=2^{(p+1)/2}$. However, $(2/q)=-1$ as $q\equiv 5\pmod{8}$. Likewise, if $i=2p$ then setting $2\equiv g^j\pmod{q}$, where $g$ is a primitive root, we get that $4p\mid 2pj$, so $j$ is even. This would again mean $2$ is a quadratic residue modulo $p$, a contradiction.

grupyorum wrote: I claim the answer is in fact $q-1=4p$. Note that $2^{4p}\equiv 1\pmod{q}$, and $i\mid 4p$ (a well-known fact). Clearly $p>2$ and $q\equiv 5\pmod{8}$. As $i\mid 4p$, we are left with $i\in\{1,2,4,p,2p,4p\}$. The cases $i=1,2,4$ are impossible since $p\ge 13$. Now if $i=p$ then $2^p\equiv 1\pmod{q}$, so $x^2\equiv 2\pmod{q}$ for $x=2^{(p+1)/2}$. However, $(2/q)=-1$ as $q\equiv 5\pmod{8}$. Likewise, if $i=2p$ then setting $2\equiv g^j\pmod{q}$, where $g$ is a primitive root, we get that $4p\mid 2pj$, so $j$ is even. This would again mean $2$ is a quadratic residue modulo $p$, a contradiction. Why $p\geq 13$, if $p=3\implies q=13$?