Multiplying with 4 (I first completed the square and then multiplied to get rid of fractions), the equation becomes
$$(2y+1)^2-(2x+21)^2+456=0$$
Thus
$$(2x+21)^2-(2y+1)^2=456$$
Factor the LHS
$$(2x-2y+20)(2x+2y+22)=456\\
(x-y+10)(x+y+11)=114$$
Now factor $114=2*3*19$ in all the $24$ possible ways each of them gives you exactly one solution.
P.S. The solution $x=19, y=27$ corresponds to
$$2 * 57=114$$
The solution $x=19, y=-28$ corresponds to
$$57 * 2=114$$
P.S. In general, by this method any equation of the form
\begin{align*}
y^2+ay-x^2-bx+c=0
\end{align*}
Can be reduced to an equation of teh form
\begin{align*}
(2y+a)^2-(2x+b)^2=\alpha
\end{align*}
which by factoring the LHS leads to finitely many solutions.
Equations of the form
\begin{align*}
y^2+ay-dx^2-bx+c=0
\end{align*}
can by multiplication by $4d^2$ be reduced to the Pell equation.
If you want the set of all solutions of this problem, then all you need to do, is the following:
Note that $2x+3y+4z=5 \iff 4(z+2x) + 3(y-2x) = 5$. Now, the general solution for an equation of this form is that $(z+2x,y-2x) = (-3n-1,4n+3)$, using the ordinary technique you have for two variables.
Hence, the final solution is $(x,y,z) = (k, 4n+2k+3,-3n-2k-1)$, where $n,k$ can vary among the integers.
For example, $n=13,k=-12$ gives $x = -12, y= 31, z = -16$, and $2x+3y+4z = 5$.
What helped here is the technique of reducing variables. Hence, we get the result.
Best Answer
Regard it as a equation in $x$, and rewrite: $x^2 - yx + 6y^2 - 47 = 0\implies \triangle = y^2 - 4(6y^2-47) = 188 - 23y^2\ge 0\implies y^2 \le 8\implies |y| = 0,1,2$ . And none of them yield a perfect square for $\triangle$. Thus no integer solutions !