Both equations are solvable in integers $(x,y)$ for infinitely many values of $a$. Spesifically, the first equation is solvable for every $a$ of the form $a=-(2^n+3)$ for integer $n$, while the second one for every $a=(-2)^n-1$.

This can be seen using the following easy observation. For any integer $a$, and any even postive integer $k$, if $(x,y)$ is a solution to the equation $$ 2 x^2+a x y+y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y+Y^2+k/2 = 0. $$ Indeed, $$ 2 X^2+a X Y+Y^2+k/2 = 2\left(\frac{x^2+k/2}{y}\right)^2+a\left(\frac{x^2+k/2}{y}\right)+x^2+k/2 = $$ $$ \frac{x^2+k/2}{y^2}(2(x^2+k/2)+axy+y^2) = \frac{x^2+k/2}{y^2}\cdot 0 = 0. $$ Now, starting with $k=2^n$ and applying this transformation $n$ times, we conclude that if the equation $$ 2x^2 + axy + y^2 + 2^n = 0 \quad \quad (1) $$ is solvable in $(x,y)$, then so is the equation $$ 2x^2 + axy + y^2 + 1 = 0 $$ with the same parameter $a$. It is left to note that (1) has a solution $(x,y)=(1,1)$ if $a=-(2^n+3)$.

Similarly, if for even integer $k$ pair $(x,y)$ is a solution to the equation $$ 2 x^2+a x y-y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y-Y^2-k/2 = 0. $$ Starint with equation $$ 2x^2 + axy - y^2 - (-2)^n = 0 \quad \quad (2) $$ with solution $(x,y)$ and applying this n times, we conclude that equation $$ 2x^2 + axy - y^2 - 1 = 0 $$ is solvable in $(x,y)$ for the same $a$. But (2) has a solution $(x,y)=(1,1)$ for $a=(-2)^n-1$.

Both equations are solvable in integers $(x,y)$ for infinitely many values of $a$. Specifically, the first equation is solvable for every $a$ of the form $a=-(2^n+3)$ for integer $n$, while the second one for every $a=(-2)^n-1$.

This can be seen using the following easy observation. For any integer $a$, and any even positive integer $k$, if $(x,y)$ is a solution to the equation $$ 2 x^2+a x y+y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y+Y^2+k/2 = 0. $$ Indeed, $$ 2 X^2+a X Y+Y^2+k/2 = 2\left(\frac{x^2+k/2}{y}\right)^2+a\left(\frac{x^2+k/2}{y}\right)+x^2+k/2 = $$ $$ \frac{x^2+k/2}{y^2}(2(x^2+k/2)+axy+y^2) = \frac{x^2+k/2}{y^2}\cdot 0 = 0. $$ Now, starting with $k=2^n$ and applying this transformation $n$ times, we conclude that if the equation $$ 2x^2 + axy + y^2 + 2^n = 0 \quad \quad (1) $$ is solvable in $(x,y)$, then so is the equation $$ 2x^2 + axy + y^2 + 1 = 0 $$ with the same parameter $a$. It is left to note that (1) has a solution $(x,y)=(1,1)$ if $a=-(2^n+3)$.

Similarly, if for even integer $k$ pair $(x,y)$ is a solution to the equation $$ 2 x^2+a x y-y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y-Y^2-k/2 = 0. $$ Starting with equation $$ 2x^2 + axy - y^2 - (-2)^n = 0 \quad \quad (2) $$ with solution $(x,y)$ and applying this n times, we conclude that equation $$ 2x^2 + axy - y^2 - 1 = 0 $$ is solvable in $(x,y)$ for the same $a$. But (2) has a solution $(x,y)=(1,1)$ for $a=(-2)^n-1$.