No need to use overcomplicated ideas. (Hopefully I have not made a mistake in these rushed calculations.) Clearly $k=3x$ and then $\ell = 4x-4$, so the congruence becomes $37x^2 - 44x + 13 \equiv 0 \pmod p$. For $p=3$ we get $(x-1)^2 \equiv 0 \pmod 3$ and $k$ and $\ell$ are both divisible by $3$. For $p=5$ we get $2x^2 - 4x + 3 \equiv 0 \pmod 5$, impossible by direct check. For $p=7$ we get $2x^2 - 2x + 6 \equiv 0 \pmod 7$, again impossible by direct check. For $p=11$ we get $4x^2 + 2 \equiv 0 \pmod {11}$, with $x\equiv 4 \pmod {11}$ as a solution. To ensure that $k$ and $\ell$ are coprime just pick $x=59$ (hmm even $x=15$ works actually).