If there are no solutions, the number of solutions is $0$. Now assume that $(u,v)$ is a solution, so $u^2-3v^2+2=16m$ and $u-1 \ge 2v$. I claim that $(2u-3v,u-2v)$ is a solution as well. This follows from that $2u-3v \ge 1$, $u-2v \ge 1$, $(2u-3v)^2-3(u-2v)^2+2=u^2-3v^2+2=16m$, $2u-3v-1 \ge 2(u-2v)$. Now this pairs $(u,v)$ and $(2u-3v,u-2v)$ together. Note that $2(2u-3v)-3(u-2v)=u$ and $(2u-3v)-2(u-2v)=v$. This implies that using this map again gives the original solution back. Now we have to check the case where $u=2u-3v$, $v=u-2v$, i.e. $u=3v$. Plugging this in, we have $6v^2+2=16m$, but this forces $v^2 \equiv 5 \pmod{8}$, which is impossible. Therefore, it is possible to pair a solution to another solution. This gives us that the number of solutions is even, as desired $\blacksquare$

I took notice that if $m=5$, there are $(9,1),(15,7)$ and if $m=3$, there are $(7,1),(11,5)$. Then I noticed that $7-1=11-5$ and $9-1=15-7$. Let $x^2-3y^2=u^2-3v^2$. We let $x-u=y-v$. Now this also gives us $x+u=3y+3v$. Solving for $x,y$ gives $x=2u-3v$ and $y=u-2v$. The solution follows.

From $x^2-3y^2+2 \equiv x^2+y^2+2 \equiv 0 \pmod{4}$, we have $x \equiv y \equiv 1 \pmod{2}$. Let $x=2a-1$ and $y=2b-1$. From $x-1 \ge 2y$, we have $a \ge 2b$. Let $a=2b+c-1$. Expansion gives $b^2-3b+c^2-3c+4bc+2=4m$. Since this expression is symmetric wrt $b,c$, we only have to check the case $b=c$. However, if $b=c$, we have $2 \equiv 2(b^2-3b)+4b^2+2 = 4m \equiv 0 \pmod{4}$, a contradiction. Therefore, the number of solutions must be even. $\blacksquare$

Expansion gives $b^2-2b+c^2-3c+4bc+2=4m$. Since this expression is symmetric wrt $b,c$, we only have to check the case $b=c$.

If there are no solutions, the number of solutions is $0$. Now assume that $(u,v)$ is a solution, so $u^2-3v^2+2=16m$ and $u-1 \ge 2v$. I claim that $(2u-3v,u-2v)$ is a solution as well. This follows from that $2u-3v \ge 1$, $u-2v \ge 1$, $(2u-3v)^2-3(u-2v)^2+2=u^2-3v^2+2=16m$, $2u-3v-1 \ge 2(u-2v)$. Now this pairs $(u,v)$ and $(2u-3v,u-2v)$ together. Note that $2(2u-3v)-3(u-2v)=u$ and $(2u-3v)-2(u-2v)=v$. This implies that using this map again gives the original solution back. Now we have to check the case where $u=2u-3v$, $v=u-2v$, i.e. $u=3v$. Plugging this in, we have $6v^2+2=16m$, but this forces $v^2 \equiv 5 \pmod{8}$, which is impossible. Therefore, it is possible to pair a solution to another solution. This gives us that the number of solutions is even, as desired $\blacksquare$

I took notice that if $m=5$, there are $(9,1),(15,7)$ and if $m=3$, there are $(7,1),(11,5)$. Then I noticed that $7-1=11-5$ and $9-1=15-7$. Let $x^2-3y^2=u^2-3v^2$. We let $x-u=y-v$. Now this also gives us $x+u=3y+3v$. Solving for $x,y$ gives $x=2u-3v$ and $y=u-2v$. The solution follows.