14. Suppose the original equation has integer solutions $(x, y)$. If $11 \mid y$, then $x^{2} \equiv 7(\bmod 11)$; if $11 \nmid y$, then by Fermat's Little Theorem, we know $11 \mid\left(y^{10}-1\right)$, i.e., $11 \mid\left(y^{5}-1\right)\left(y^{5}+1\right)$. Since $\left(y^{5}-1, y^{5}+1\right) \mid 2$, it follows that $y^{5} \equiv \pm 1(\bmod 11)$, hence $x^{2} \equiv 6$ or $8(\bmod 11)$. Therefore, we need $x^{2} \equiv 6, 7$ or $8(\bmod 11)$. However, by discussing $x \equiv 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$ $(\bmod 11)$, we find that $x^{2} \equiv 0, 1, 4, 9, 5$ or $3(\bmod 11)$, and it is impossible for $x^{2} \equiv 6$, 7 or $8(\bmod 11)$, leading to a contradiction.