Anyone ideas? bump

No. Suppose such an integer exists. We assume it is of the form \(c^k=\overline{2\_\_\dots\_\_2}\). If it ends with \(n\) zeroes, then \(2\mid c\) and \(5\mid c\) so \(10\mid c\). Let \(c=10^ab\) where \(10\nmid b\). Then \(c^k=10^{ak}b^k\), and \(ak=n\). So \(c^k/10^n=b^k\) is also a perfect power with only \(2\)s and \(0\)s, which ends in \(2\). But then since \(2\mid c\) and \(k\geq2\), we have \(4\mid c\). So the last two digits must be divisible by \(4\). But they can be only \(02\) or \(22\), neither of which is divisible by \(4\). So this is impossible.