Partial sol. $2^n+15=k(3x^2-4xy+3y^2)$ Case $k=1$ Let $u=x+y, v=x-y$, the $RHS$ becomes $\frac{u^2+5v^2}{2}\Longrightarrow$ $$2^N+30=u^2+5v^2$$wherein $N=n+1$ Taking mod $5\Longrightarrow 2^N=u^2\Longrightarrow 2^N\equiv 2,4,3,1\bmod{5}$ (cyclic), $u^2\equiv1,4\bmod{5}$ Casework $2^4, 2^6\equiv 4\bmod {5,}\Rightarrow u=5k+2$ $k=0\Longrightarrow$ No solutions $k=1\Longrightarrow u=7, v=3 \Rightarrow 94=2^6+30=7^2+5.3^2=49+45 \Rightarrow(u,v,N)=(7,3,6)$ $\Rightarrow (x,y, n)=(5,2,5), (2,5,5), (-5,-2,5),(-2,-5,5)$. Therefore lowest number looked for $=47$ Case $k>1$ There is apparently only one solution $2^{14}+30=29(21^2+5.5^2)$ with higher $(u,v,N)$.

$d=3x^2-4xy+3y^2=2(x-y)^2+x^2+y^2$. $d|2^n+15$ (odd number) $\Longrightarrow d$ is odd number $\Longrightarrow x,y$ have different parities $\Longrightarrow x-y$ is an odd number. WLOG, we can assume $x>y$. Case 1: $x-y=1$. We obtain the smallest possible values of $d$ for the pairs $(x,y)\in\{(1,0),(2,1),(3,2),(4,3)\}$: $d\in\{3,7,15,27\}$. We can verify: $3\nmid 2^n+15;\;7\nmid 2^n+15;\;15\nmid 2^n+15;\;27\nmid 2^n+15,\;\forall n\in\mathbb{N}$. Case 2: $x-y=3$. We obtain the smallest possible value of $d$ for the pair $(x,y)=(2,-1)$: $d=23=2^3+15$. Case 3: $x-y\ge5$. $d>2\cdot5^2=50$. Hence, the smallest requested number is $d=23$, obtained for $n=3,\;x=2,\;y=-1$.

Solution via quadratic reciprocity. Set $k\triangleq 3x^2 - 4xy + 3y^2$. Let $u=x+y$ and $v=x-y$. Then, $k=(u^2+5v^2)/2$. Take any prime $p\mid k$. Since $(k,30)=1$, it follows $p\ge 7$. In particular, if $p\mid u,v$ then $k\ge 3p^2\ge 147$. Assume in the remainder $p\nmid u,v$. We then have $(-5/p)=1$. Case 1. $p\equiv 1\pmod{4}$. In this case, $(-1/p)=1$, hence $(5/p)=1$. By quadratic reciprocity, $(5/p)(p/5)=1$, yielding $(p/5)=1$. Since $p\ne 5$ (see above), it follows $p\equiv \pm 1\pmod{4}$. Consequently, either $p\equiv 1\pmod{20}$ or $p\equiv 9\pmod{20}$. Case 2. $p\equiv 3\pmod{4}$. In this case, $(-1/p)=-1$, hence $(5/p)=-1$. Once again, $(5/p)(p/5)=1$ by quadratic reciprocity. Thus, $(p/5)=-1$. Hence, $p\equiv \pm 2\pmod{5}$. This yields either $p\equiv 7\pmod{20}$ or $p\equiv 3\pmod{20}$. Combining Case 1 and Case 2, we find $p\in\{1,3,7,9\}\pmod{20}$ for any prime $p$ dividing $k=3x^2-4xy+3y^2=(u^2+5v^2)/2$. Clearly, the smallest such prime $p$ is $23$, hence $k\ge 23$. We now find an example. We have $u^2+5v^2=46$ which is enjoyed for $(u,v) = (1,3)$ yielding $(x,y)=(2,-1)$. Finally, for $(x,y)=(2,-1)$ (thus $k=23$) and $n\equiv 0\pmod{3}$, we indeed have $k\mid 2^n+15$, concluding the proof.

Let $3x^2-4xy+3y^2=k \rightarrow 3x^2-4xy+3y^2-k=0$. $x=\frac{2y+\sqrt{3k-5y^2}}{3}$ or $x=\frac{2y-\sqrt{3k-5y^2}}{3}$ We know that $2\nmid 2^n+15 , 3\nmid 2^n+15 , 5\nmid 2^n+15$ and $3x^2-4xy+3y^2\neq 1$ The smallest possible number is $7$. Check for $k=7$, we have $x=\frac{2y+\sqrt{21-5y^2}}{3}$ or $x=\frac{2y-\sqrt{21-5y^2}}{3}$ Set $y=2 , x=\frac{4-\sqrt{21-20}}{3} \iff x=1$ but $7\nmid 2^n+15$ Check $k=11 \rightarrow 33-5y^2=m^2 \rightarrow m^2\equiv 3(mod 5)$ contradiction. Check $k=13 \rightarrow 39-5y^2=x^2$ no solution. Check $k=17$ no solution Check $k=19$ no solution Check $k=23 \rightarrow x=\frac{2y+\sqrt{69-5y^2}}{3}$ or $x=\frac{2y-\sqrt{69-5y^2}}{3}$ Set $y=1 , x=\frac{2-\sqrt{69-5}}{3} \iff x=-2$ $23|2^n+15$ one of solution $n=3$ the answer is $23$

No need to do crazy things. Note $3x^2 - 4xy + 3y^2 = 3(x - \frac{2y}{3})^2 + \frac{5y^2}{3}$ and that $2^n + 15$ is not divisible by $2$, $3$, $5$ (clearly) and $7$ (since the remainders of powers of $2$ are $2$, $4$ and $1$). Since $x = 2, y = -1$ gives $23 = 2^3 + 15$, it remains to prove no smaller ones are possible. Having in mind the above divisibility restriction, we are left to check the numbers $1$, $11$, $13$, $17$ and $19$. In any case we cannot have $|y| \geq 4$, as then $\frac{5y^2}{3} > 19$. For $y= 0, \pm 3, \pm 2, \pm 1$ we get the quadratics $3x^2$, $3x^2 \pm 12x + 27$ (all up until now are divisible by $3$, so impossible), $3x^2 \pm 8x + 12$ and $3x^2 \pm 4x + 3$. Solving the corresponding quadratic equation shows that the latter are not equal to $1$, $11$, $13$, $17$, $19$ for any integer $x$.