Here is a conditional answer. It was shown by Macintyre and Wilkie that if (a weak variant of) Schanuel's Conjecture is true, then the first-order theory of the real exponential field $(\mathbb{R};0,1,+,\times,\exp)$ is decidable. In particular, the (very unwieldy) first-order sentence $$\bigvee_{n=A}^B \exp\exp\exp 79 = n,$$ where $A = 2^{2^{2^{79}}}$ and $B = 3^{3^{3^{79}}}$ is then decidable by finitary means. Of course, it is conceivable that Schanuel's Conjecture is false...
As pointed out by Dave Marker and George Lowther, Schanuel's Conjecture directly implies that $e^{e^{e^{79}}}$ is not an integer. Indeed, since $e^{79}$ is irrational, $79$ and $e^{79}$ are linearly independent over $\mathbb{Q}$, which means that $e^{79},e^{e^{79}}$ are algebraically independent over $\mathbb{Q}$. Since $79,e^{79},e^{e^{79}}$ are therefore linearly independent over $\mathbb{Q}$, it follows that $e^{79},e^{e^{79}},e^{e^{e^{79}}}$ are algebraically independent over $\mathbb{Q}$. In particular, $e^{e^{e^{79}}}$ is not an integer.